Solve Quadratic Equation: x²-3x-5=10+3x+12

Let's notice that this is a quadratic equation, therefore we will first arrange it and move all terms to one side and 0 to the other side:

$x^2-3x-5=10+3x+12 \\ x^2-3x-3x-5-10-12=0\\ x^2-6x-27=0$

where in the first stage we moved all terms to one side while remembering that when moving a term its sign changes and in the second stage we combined like terms,

Now we want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this is not possible, since there is no factor that is common to all three terms on the left side of the equation, therefore, we'll look for trinomial factoring:

Let's note that the coefficient of the quadratic term (the term with power of 2) is 1, therefore we can try to use quick trinomial method: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial method:

The rule states that for the quadratic algebraic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$that satisfy the following conditions (quick trinomial method conditions):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation we got in the last stage after arranging it:

$x^2-6x-27=0$

Let's note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=-27 \\ b=-6 \end{cases}$

where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial method above, so we'll look for two numbers $m,\hspace{4pt}n$that satisfy:

$\begin{cases} m\cdot n=-27 \\ m+n=-6 \end{cases}$

Let's try to find these numbers through logical thinking and using our multiplication table knowledge, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we mentioned in the last stage:

$m\cdot n=-27$

We can identify that their product must give a negative result, therefore we can conclude that their signs are different,

Next we'll look at the factors (whole numbers) of 27, and from our multiplication table knowledge we know there are two possibilities for such factors: 3 and 9, 27 and 1, where we previously concluded their signs must be different, a quick check of these possibilities regarding the second condition (and there's always an obligation to check condition satisfaction, even if there's only one possible pair of whole number factors as mentioned above):

$m+n=-6$

will lead to the quick conclusion that the only possibility for satisfying both conditions above together is:

$-9,\hspace{4pt}3$meaning - for:

$m=-9,\hspace{4pt}n=3$

(it doesn't matter which we call m and which we call n)

It satisfies that:

$\begin{cases} (\underline{-9})\cdot\underline{3}=-27 \\ (\underline{-9})+\underline{3}=-6 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2-6x-27\\ \downarrow\\ (x+(-9))(x+3)$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial method, and the equation is:

$x^2-6x-27=0 \\ \downarrow\\ (x+(-9))(x+3)=0\\ (x-9)(x+3)=0\\$where in the second stage we just simplified the expression inside the left parentheses in the above product.

Now that the expression on the left side is factored let's continue to its quick solution,

Let's note a simple fact, on the left side there's a product of two terms, and on the right side the number 0,

Therefore we can conclude that the only two possibilities for which this equation will be satisfied are if:

$x-9=0$

or if:

$x+3=0$

since only multiplying a number by 0 will give the result 0,

From here we'll solve the two new equations we got:

$x-9=0 \rightarrow x=9\\ x+3=0\rightarrow x=-3$

where we solved each equation separately,

Let's summarize: We therefore got the solutions of the quadratic equation and used quick trinomial method to factor the quadratic expression on its left side:

$x^2-6x-27=0 \\ \downarrow\\ (x-9)(x+3)=0\\$which are:

$x_1=9,\hspace{4pt}x_2=-3$

where substituting either of these solutions, the first or the second, in the equation - will give a true statement,

Therefore the correct answer is answer B.