The Quadratic Formula: Complete the one solution equation

To solve the quadratic equation $X^2 + 4X + 4 = 0$, we can attempt to factor it as a perfect square trinomial.

• Step 1: Recognize that the equation is a perfect square trinomial because it can be written as $(X + 2)^2$. The relationship is $(X + 2)(X + 2) = X^2 + 4X + 4$.
• Step 2: Set the factored equation equal to zero: $(X + 2)^2 = 0$.
• Step 3: Solve for $X$ by setting $X + 2 = 0$.
• Step 4: Solving $X + 2 = 0$ gives $X = -2$.

Therefore, the solution to the equation is $X = -2$.

Thus, the correct answer choice is $x = -2$, corresponding to choice 1.

To solve the quadratic equation $4x^2 + 24x + 36 = 0$, we will simplify it by factoring:

First, notice that the given equation can be simplified as a perfect square:

• Recognize that $4x^2$, $24x$, and $36$ can form a perfect square trinomial: $(2x + 6)^2$.
• Expand $(2x + 6)^2$ to verify it corresponds to the original equation:
  $(2x + 6)^2 = (2x + 6)(2x + 6) = 4x^2 + 12x + 12x + 36 = 4x^2 + 24x + 36$.

The equation has now been verified to be a perfect square: $\left(2x + 6\right)^2 = 0$.

Set $2x + 6 = 0$, and solve for $x$:

• Subtract 6 from both sides: $2x = -6$.
• Divide both sides by 2: $x = -3$.

Thus, the solution to the quadratic equation is $\boxed{x = -3}$.

To solve the quadratic equation $36x^2 - 144x + 144 = 0$, we first examine its structure to determine the best method for solution:

Step 1: Simplify the equation.
Notice that each term in the equation $36x^2 - 144x + 144$ is divisible by 36. Let's simplify it by dividing each term by 36:

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

Step 2: Factor the simplified equation.
The equation $x^2 - 4x + 4$ can be factored as $(x - 2)^2 = 0$, since both 2 and -2 added yield -4, and multiplied give 4.

Step 3: Solve for x.
Given $(x - 2)^2 = 0$, the solution is $x - 2 = 0$, which results in:

$x = 2$

Therefore, the solution to the equation is $x = 2$.

This corresponds to the provided correct answer choice $x=2$.

To solve this problem, we'll follow these steps:

• Step 1: Identify the coefficients $a = -81$, $b = 54$, and $c = -9$.
• Step 2: Calculate the discriminant $D = b^2 - 4ac$.
• Step 3: Apply the quadratic formula to find the solution for $x$.

Step 1: We have $a = -81$, $b = 54$, and $c = -9$.

Step 2: Calculate the discriminant:

$D = b^2 - 4ac = 54^2 - 4(-81)(-9)$

$= 2916 - 2916 = 0$

Since the discriminant is zero, there is exactly one real solution, indicating a perfect square trinomial.

Step 3: Apply the quadratic formula:

$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-54 \pm \sqrt{0}}{-162}$

$x = \frac{-54}{-162} = \frac{1}{3}$

Therefore, the solution to the problem is $x = \frac{1}{3}$.

To solve this quadratic equation $-4x^2 + 96x - 576 = 0$, we will apply the quadratic formula:

• The general form of a quadratic equation is $ax^2 + bx + c = 0$.
• For this equation, the coefficients are $a = -4$, $b = 96$, and $c = -576$.
• The quadratic formula states: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• First, compute the discriminant: $b^2 - 4ac = 96^2 - 4(-4)(-576)$.
• Calculate $96^2 = 9216$ and $4(-4)(-576) = 9216$ as well.
• Thus, the discriminant is $9216 - 9216 = 0$.
• A discriminant of zero indicates that there is exactly one real solution (a repeated root).
• Now substitute into the quadratic formula: $x = \frac{-96 \pm \sqrt{0}}{2(-4)}$.
• This simplifies to $x = \frac{-96}{-8}$.
• Further simplification gives $x = 12$.

Therefore, the solution to the equation is $x = 12$, which corresponds to answer choice 2.

To solve the quadratic equation $81x^2 - 216x + 144 = 0$, we begin by identifying the coefficients: $a = 81$, $b = -216$, and $c = 144$.

Next, we use the quadratic formula:

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

Substitute the values of $a$, $b$, and $c$ into the formula:

$b^2 - 4ac = (-216)^2 - 4 \cdot 81 \cdot 144$

$= 46656 - 46656 = 0$

The discriminant is zero, indicating there is exactly one real solution. Substitute back into the quadratic formula:

$x = \frac{-(-216) \pm \sqrt{0}}{2 \cdot 81}$

$x = \frac{216 \pm 0}{162}$

$x = \frac{216}{162}$

Simplify the fraction:

$x = \frac{216 \div 54}{162 \div 54} = \frac{4}{3}$

Therefore, the solution to the equation is $x = \frac{4}{3}$.

By comparing with the given choices, choice 1: $x = \frac{4}{3}$ is correct.

To solve the quadratic equation $9y^2 - 30y = -25$, we first rewrite it in standard form:

$9y^2 - 30y + 25 = 0$.

Identifying the coefficients, we have $a = 9$, $b = -30$, and $c = 25$. We will apply the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

First, compute the discriminant:

$b^2 - 4ac = (-30)^2 - 4 \cdot 9 \cdot 25 = 900 - 900 = 0$.

Since the discriminant is zero, there is a single repeated root. Substituting back into the quadratic formula, we get:

$y = \frac{-(-30) \pm \sqrt{0}}{2 \cdot 9} = \frac{30}{18} = \frac{5}{3}$.

Therefore, the solution to the equation is $y = \frac{5}{3}$.

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.

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.

To solve the equation $\frac{x^2}{4} + x + 1 = 0$, we first rewrite it in the standard quadratic form:

$\frac{x^2}{4} + x + 1 = 0$ becomes $\frac{1}{4}x^2 + x + 1 = 0$.

Identifying the coefficients, we have:

• $a = \frac{1}{4}$
• $b = 1$
• $c = 1$

Next, we use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Plugging in the coefficients, we get:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot \frac{1}{4} \cdot 1}}{2 \cdot \frac{1}{4}}$.

Calculate the discriminant:

$b^2 - 4ac = 1 - 1 = 0$.

Since the discriminant is zero, there is exactly one real root. Substitute back into the quadratic formula:

$x = \frac{-1 \pm \sqrt{0}}{\frac{1}{2}}$.

$x = \frac{-1}{\frac{1}{2}}$.

$x = -2$.

Therefore, the solution to the equation is $x = -2$, which corresponds to choice 2.

To solve this equation, we shall proceed with these steps:

• Step 1: Clear the fractions. Multiply each term of the equation by 9 to simplify:
• $9 \left(\frac{x^2}{9}\right) + 9 \left(\frac{2}{9}x\right) + 9 \left(\frac{1}{9}\right) = 9 \times 0$
• This simplifies down to $x^2 + 2x + 1 = 0$.
• Step 2: Recognize $x^2 + 2x + 1 = 0$ as a quadratic equation, which can be factored as:
• $(x + 1)^2 = 0$.
• Step 3: Solving $(x + 1)^2 = 0$ gives $x + 1 = 0$, thus $x = -1$.

Therefore, the solution to the equation is $x = -1$.

The correct choice that corresponds to this solution from the provided options is $x = -1$.

To solve the given quadratic equation $\frac{4}{9}x^2+\frac{8}{3}x+4=0$, we will apply the Quadratic Formula. Let's outline the steps.

• Step 1: Identify the coefficients.
  $a = \frac{4}{9}, \, b = \frac{8}{3}, \, c = 4$.
• Step 2: Calculate the discriminant.
  $\Delta = b^2 - 4ac = \left(\frac{8}{3}\right)^2 - 4\left(\frac{4}{9}\right)(4)$.

Calculating $\Delta$:
$\left(\frac{8}{3}\right)^2 = \frac{64}{9}$
$4 \times \frac{4}{9} \times 4 = \frac{64}{9}$,
so, $\Delta = \frac{64}{9} - \frac{64}{9} = 0$.

• Step 3: Use the Quadratic Formula. Since the discriminant is zero, there's a single solution.
• The formula simplifies to $x = \frac{-b}{2a}$.
• Calculate $x = \frac{-\frac{8}{3}}{2 \times \frac{4}{9}} = \frac{-\frac{8}{3}}{\frac{8}{9}} = -3$.

Therefore, the solution to the equation is $x = -3$.

To solve the equation $\frac{x^2}{4}+\frac{x}{2}+\frac{1}{4}=0$, we will follow these steps:

• Convert the equation to standard quadratic form.
• Identify the coefficients $a$, $b$, and $c$.
• Apply the quadratic formula.
• Calculate the discriminant and solve for $x$.

Step 1: The given equation is:
$\frac{x^2}{4} + \frac{x}{2} + \frac{1}{4} = 0$.

To convert it into standard form $ax^2 + bx + c = 0$, multiply the entire equation by 4 to eliminate the denominators:

$x^2 + 2x + 1 = 0$.

Step 2: Identify the coefficients:

• $a = 1$
• $b = 2$
• $c = 1$

Step 3: Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 4: Calculate the discriminant $b^2 - 4ac$:

$b^2 - 4ac = (2)^2 - 4(1)(1) = 4 - 4 = 0$.

Since the discriminant is 0, we have one real repeated solution.

Step 5: Solve for $x$:

$x = \frac{-2 \pm \sqrt{0}}{2(1)} = \frac{-2}{2} = -1$.

Therefore, the solution to the equation is $x = -1$.

To solve the given quadratic equation $x^2 + \frac{10}{9}x + \frac{25}{81} = 0$, we will first check if it can be expressed as a perfect square trinomial.

Notice that a quadratic equation in the form $(x + d)^2 = 0$ expands to:

$(x + d)^2 = x^2 + 2dx + d^2$.

We observe:

• $2d = \frac{10}{9}$, which gives $d = \frac{10}{18} = \frac{5}{9}$.
• $d^2 = \left(\frac{5}{9}\right)^2 = \frac{25}{81}$.

This confirms that the original equation can be rewritten as:

$(x + \frac{5}{9})^2 = 0$.

Solving $(x + \frac{5}{9})^2 = 0$ yields:

$x + \frac{5}{9} = 0$.

Subtracting $\frac{5}{9}$ from both sides, we get:

$x = -\frac{5}{9}$.

Therefore, the solution to the equation is $x = -\frac{5}{9}$, which corresponds to choice id="3".

To solve the quadratic equation $\frac{x^2}{4}+\frac{2}{3}x+\frac{4}{9}=0$, we will use the quadratic formula:

The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, we identify the coefficients:
$a = \frac{1}{4}$, $b = \frac{2}{3}$, $c = \frac{4}{9}$

Now, we calculate the discriminant:
$b^2 - 4ac = \left(\frac{2}{3}\right)^2 - 4 \cdot \frac{1}{4} \cdot \frac{4}{9}$

Calculating further:
$b^2 = \frac{4}{9}$
$4ac = 4 \cdot \frac{1}{4} \cdot \frac{4}{9} = \frac{16}{36} = \frac{4}{9}$

Hence, the discriminant:
$b^2 - 4ac = \frac{4}{9} - \frac{4}{9} = 0$

Since the discriminant is 0, there is exactly one real solution. We apply the quadratic formula:
$x = \frac{-\left(\frac{2}{3}\right) \pm \sqrt{0}}{2 \times \frac{1}{4}}$

Simplify:
$x = \frac{-\frac{2}{3}}{\frac{1}{2}} = -\frac{2}{3} \cdot 2 = -\frac{4}{3}$

Therefore, the solution to the problem is $x = -\frac{4}{3}$.

To solve the quadratic equation $x^2 + x + \frac{1}{4} = 0$, we will use the method of completing the square:

• Step 1: Rewrite the equation focusing on forming a perfect square.
• Step 2: The expression $x^2 + x$ can be transformed into a perfect square.
• Step 3: Note that $(x + \frac{1}{2})^2 = x^2 + x + \frac{1}{4}$.
• Step 4: Substitute into the equation: $(x + \frac{1}{2})^2 = 0$.
• Step 5: Solve for $x$: If $(x + \frac{1}{2})^2 = 0$, then $x + \frac{1}{2} = 0$.
• Step 6: Simplifying the equation, we find $x = -\frac{1}{2}$.

This means the solution to the quadratic equation is $x = -\frac{1}{2}$.

Thus, the correct answer is $x = -\frac{1}{2}$.

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:

$y^2+4y+2=-2$

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

$y^2+4y+2=-2 \\ y^2+4y+2+2=0 \\ y^2+4y+4=0$

Note that the expression on the left side can be factored 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 shown below:

$4=2^2$

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

$y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+4y+\textcolor{blue}{2}^2=0$

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:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2=0$

Note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{2}^2$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 the expression in question (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 ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2+\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}}+\textcolor{blue}{2}^2 =0$

And indeed it is true that:

$2\cdot y\cdot2=4y$

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

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

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:

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

Let's summarize the solution of the equation:

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

Therefore the correct answer is answer D.

To solve the given equation, follow these steps:

• Step 1: Expand $(x - 4)^2$ using the formula $(a - b)^2 = a^2 - 2ab + b^2$.

Thus, $(x - 4)^2 = x^2 - 8x + 16$.

• Step 2: Substitute the expanded form into the equation:

$x^2 - 8x + 16 + 3x^2 = -16x + 12$.

• Step 3: Combine like terms on the left-hand side.

This gives $4x^2 - 8x + 16 = -16x + 12$.

• Step 4: Rearrange the equation to set it to zero.

Bring all terms to one side: $4x^2 - 8x + 16 + 16x - 12 = 0$.

Combine and simplify the terms: $4x^2 + 8x + 4 = 0$.

• Step 5: Simplify the equation by dividing each term by 4.

It becomes $x^2 + 2x + 1 = 0$.

• Step 6: Recognize the equation as a perfect square trinomial.

$(x + 1)^2 = 0$.

• Step 7: Solve by taking the square root of both sides.

The solution is $x + 1 = 0$, therefore $x = -1$.

In conclusion, the solution to the equation is $x = -1$.

To solve the equation $(x+3)^2 = 2x + 5$, we proceed as follows:

• Step 1: Expand the left side. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we find:
  $(x+3)^2 = x^2 + 6x + 9$.

• Step 2: Set the equation to zero by moving all terms to one side:
  $x^2 + 6x + 9 = 2x + 5$
  Subtract $2x + 5$ from both sides:
  $x^2 + 6x + 9 - 2x - 5 = 0$
  This simplifies to:
  $x^2 + 4x + 4 = 0$.

• Step 3: Solve the quadratic equation $x^2 + 4x + 4 = 0$. Notice this can be factored as:
  $(x+2)^2 = 0$.

• Step 4: Solve for $x$ by setting the factor equal to zero:
  $x+2 = 0$.
  Thus, $x = -2$.

Therefore, the solution to the equation $(x+3)^2 = 2x + 5$ is $x = -2$.