Solving Quadratic Equations - Examples, Exercises and Solutions

Let's recall the general form of the quadratic function:

$y=ax^2+bx+c$ The function given in the problem is:

$y=-x^2+25x$$c$is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

$c=0$Therefore, the correct answer is answer A.

This is a quadratic equation:

$x^2+5x+4=0 =0$

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in to a form where all terms on one side are ordered from the highest to the lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula.

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+5x+4=0 =0$

And solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=5 \\ c=4\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the solutions of the equation (its roots) by insertion we just identified into the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot4}}{2\cdot1}$

Let's continue to calculate the expression inside of the square root and simplify the expression:

$x_{1,2}=\frac{-5\pm\sqrt{9}}{2}=\frac{-5\pm3}{2}$

Therefore the solutions of the equation are:

$$$\begin{cases}x_1=\frac{-5+3}{2}=-1 \\ x_2=\frac{-5-3}{2}=-4\end{cases}$

Therefore the correct answer is answer C.

This is a quadratic equation:

$x^2+5x+6=0$

due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it to a form where all the terms on one side are ordered from the highest to the lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+5x+6=0$and solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=5 \\ c=6\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the equation's solutions (roots) by inserting the coefficients we just noted into the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}$

Let's continue to calculate the expression inside of the square root and proceed to simplify the expression:

$x_{1,2}=\frac{-5\pm\sqrt{1}}{2}=\frac{-5\pm1}{2}$

The solutions to the equation are:

$$$\begin{cases}x_1=\frac{-5+1}{2}=-2 \\ x_2=\frac{-5-1}{2}=-3\end{cases}$

Therefore the correct answer is answer D.

This is a quadratic equation:

$x^2+3x-18=0$

This is due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of the equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we obtain a true statement when inserted into the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2+3x-18=0$

And solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1\\b=3\\c=-18\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

And we'll obtain the solutions of the equation (its roots) by substituting the coefficients we just noted in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-18)}}{2\cdot1}$

Let's continue and calculate the expression inside the square root and simplify the expression:

$x_{1,2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}$

Therefore the solutions of the equation are:

$$$\begin{cases}x_1=\frac{-3+9}{2}=3 \\ x_2=\frac{-3-9}{2}=-6\end{cases}$

Therefore the correct answer is answer C.

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We substitute into the formula:

-5±√(5²-4*1*4) 
          2

-5±√(25-16)
         2

-5±√9
    2

-5±3
   2

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

-5-3 = -8
-8/2 = -4

-5+3 = -2
-2/2 = -1

And thus we find out that X = -1, -4

Let's recall the quadratic formula:

We'll substitute the given data into the formula:

$x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}$

Let's simplify and solve the part under the square root:

$x={{10\pm\sqrt{100+96}\over 4}}$

$x={{10\pm\sqrt{196}\over 4}}$

$x={{10\pm14\over 4}}$

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

$x={{10+14\over 4}} = {24\over4}=6$

$x={{10-14\over 4}} = {-4\over4}=-1$

We've arrived at the solution: X=6,-1

To answer the question, we'll need to recall the quadratic formula:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

Let's remember that:

a is the coefficient of X²

b is the coefficient of X

c is the free term

And if we look again at the formula given to us:

a=1

b=10

c=9

Let's substitute into the formula:

$x = {-10 \pm \sqrt{10^2-4\cdot 1 \cdot 9} \over 2\cdot 1}$

Let's start by solving what's under the square root:

$x = {-10 \pm \sqrt{100-36} \over 2}$

$x = {-10 \pm \sqrt{64} \over 2}$

$x = {-10 \pm 8 \over 2}$

Now we'll solve twice, once with plus and once with minus

$x = {-10 +8 \over 2}= {-2 \over 2} = -1$

$x = {-10 -8 \over 2} = {-18 \over 2} =-9$

And we can see that we got two solutions, X=-1 and X=-9

And that's the solution!

The following is a quadratic equation:

$2x^2-3x+5=0$

This is due to the fact that there is a quadratic term (meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed to solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation in the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$2x^2-3x+5=0$and solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=2 \\ b=-3 \\ c=5\end{cases}$

where we noted that the coefficient includes the minus sign, and this is because in the general form of the equation we mentioned earlier:

$ax^2+bx+c=0$

the coefficients are defined such that they have a plus sign in front of them, and therefore the minus sign must be included in the coefficient value.

Let's continue and obtain the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot2\cdot5}}{2\cdot2}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{3\pm\sqrt{-31}}{2}\frac{}{}$

The expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of $x$that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

This is a quadratic equation:

$-x^2+6x-10=0$

due to the fact that there is a quadratic term present(meaning raised to the second power),

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side,

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

For ease of solving and minimizing errors, it is always recommended to ensure that the coefficient of the quadratic term in the equation is positive,

We'll achieve this by multiplying (both sides of) the equation by:$-1$:

$-x^2+6x-10=0 \hspace{8pt}\text{/}\cdot(-1)\\ x^2-6x+10=0$$$

Let's continue solving the equation

Solve it using the quadratic formula,

Remember:

The rule states that the roots of an equation of the form:

$ax^2+bx+c=0$

are:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

(meaning its solutions, the two possible values of the unknown for which we get a true statement when substituted in the equation)

$$This formula is called: "The Quadratic Formula"

Let's return to the problem:

$x^2-6x+10=0$

and solve it:

First, let's identify the coefficients of the terms:

$\begin{cases}a=1 \\ b=-6 \\ c=10\end{cases}$

where we noted that the coefficient of the quadratic term is 1,

We obtain the equation's solutions (roots) by substituting these coefficients that we mentioned earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot10}}{2\cdot1}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{6\pm\sqrt{-4}}{2}$

The expression under the root is negative, and since we cannot extract a real root from a negative number, this equation has no real solutions,

Meaning - there is no real value of $x$that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We identify that we have:
a=1
b=0
c=9

We recall the root formula:

We replace according to the formula:

-0 ± √(0²-4*1*9)

           2

We will focus on the part inside the square root (also called delta)

√(0-4*1*9)

√(0-36)

√-36

It is not possible to take the square root of a negative number.

And so the question has no solution.

Proceed to solve the given equation:

$x^2+10x=-25$

First, let's arrange the equation by moving terms:

$x^2+10x=-25 \\ x^2+10x+25=0 \\$Note that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As shown below:

$25=5^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a line):

$(\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we will query whether we can represent the expression on the left side as:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it is true that:

$2\cdot x\cdot5=10x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget there are two possibilities - positive and negative when taking the square root of an even power), then we'll easily solve by isolating the variable:

$(x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}$

Let's summarize the solution of the equation:

$x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$16a^2+20a+20=-5-20a$

First, let's organize the equation by moving terms and combining like terms:

$16a^2+20a+20=-5-20a \\ 16a^2+20a+20+5+20a =0\\ 16a^2+40a+25=0$

Note that we are able to factor the expression on the left side by using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+2\textcolor{red}{x}\textcolor{blue}{y}+\textcolor{blue}{y}^2$

As shown below:

$16=4^2\\ 25=5^2$

Using the law of exponents for powers applied to products in parentheses (in reverse):

$x^ny^n=(xy)^n$

Therefore, first we'll express the outer terms as a product of squared terms:

$16a^2+40a+25=0 \\ 4^2a^2+40a+5^2=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+40a+\textcolor{blue}{5}^2=0$

Now let's examine again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

And the expression on the left side of the equation that we obtained in the last step:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0$

Note that the terms $(\textcolor{red}{4a})^2,\hspace{6pt}\textcolor{blue}{5}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{x}+\textcolor{blue}{y})^2=\textcolor{red}{x}^2+\underline{2\textcolor{red}{x}\textcolor{blue}{y}}+\textcolor{blue}{y}^2$

In other words - we are querying whether we can express the expression on the left side of the equation as:

$(\textcolor{red}{4a})^2+\underline{40a}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ (\textcolor{red}{4a})^2+\underline{2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it holds that:

$2\cdot4a\cdot5=40a$

Therefore, we can express the expression on the left side of the equation as a perfect square binomial:

$(\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable and dividing both sides of the equation by the variable's coefficient:

$(4a+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ 4a+5=\pm0\\ 4a+5=0\\ 4a=-5\hspace{8pt}\text{/}:4\\ \boxed{a=-\frac{5}{4}}$

Let's summarize the solution of the equation:

$16a^2+20a+20=-5-20a \\ 16a^2+40a+25=0 \\ \downarrow\\ (\textcolor{red}{4a})^2+2\cdot\textcolor{red}{4a}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0\\ \downarrow\\ (\textcolor{red}{4a}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ 4a+5=0\\ \downarrow\\ \boxed{a=-\frac{5}{4}}$

Therefore the correct answer is answer D.

Let's solve the given equation:

$60-16y+y^2=-4$First, let's arrange the equation by moving terms:

$60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$Now let's examine again the short factoring formula we mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$And the expression on the left side of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$And indeed it holds that:

$2\cdot y\cdot8=16y$So we can present the expression on the left side of the given equation as a difference of two squares:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

$(y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}$

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

So the correct answer is answer a.

Solve the given equation:

$x^2-10x=-16$

First, let's arrange the equation by moving terms:

$x^2-10x=-16 \\ x^2-10x+16 =0$

Note that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side by using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=16\\ m+n=-10\\$From the first requirement, namely - the multiplication, we observe that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same sign, according to multiplication rules. Remember that the possible factors of 16 are the number pairs 4 and 4, 2 and 8, or 16 and 1. Meeting the second requirement, along with the fact that the signs of the numbers we're looking for are identical leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=-2 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0$

From remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll obtain two simple equations and solve them by isolating the unknown in each:

$x-8=0\\ \boxed{x=8}$

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$x^2-10x=-16 \\ x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=8,2}$

Therefore the correct answer is answer B.

Proceed to solve the given equation:

$x^2+144=24x$

Arrange the equation by moving terms:

$x^2+144=24x \\ x^2-24x+144=0$

Note that we are able to factor the expression on the left side by using the perfect square trinomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

As demonstrated below:

$144=12^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0$

Now let's examine once again the perfect square trinomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side in the equation that we obtained in the last step:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll query whether we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot12=24x$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}$

Let's summarize the solution of the equation:

$x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}$

Therefore the correct answer is answer C.