Solve the Quadratic Equation: 3x²-12x=36 Step by Step

Let's solve the following equation:

$3x^2-12x=36$

$$Let's begin by rearranging the equation:

$3x^2-12x=36 \\ 3x^2-12x-36=0$

Note that all coefficients as well as the free term are multiples of 3, hence we'll divide both sides of the equation by 3:

$3x^2-12x-36=0 \hspace{6pt}\text{/}:3 \\ x^2-4x-12=0$

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

$m\cdot n=-12\\ m+n=-4\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers have different signs, according to multiplication rules. The possible factors of 12 are 6 and 2, 4 and 3, or 12 and 1. Meeting the second requirement mentioned, along with the fact that the numbers we're looking for have different signs lead us to the conclusion that the only possibility for the two numbers is:

$\begin{cases}m=-6 \\ n=2\end{cases}$

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

$x^2-4x-12=0 \\ \downarrow\\ (x-6)(x+2)=0$

Remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain two simple equations and solve them by isolating the variable in each:

$x-6=0\\ \boxed{x=6}$

or:

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

Let's summarize the solution of the equation:

$3x^2-12x=36 \\ x^2-4x-12=0 \\ \downarrow\\ (x-6)(x+2)=0 \\ \downarrow\\ x-6=0\rightarrow\boxed{x=6}\\ x+2=0\rightarrow\boxed{x=-2}\\ \downarrow\\ \boxed{x=6,-2}$

Therefore the correct answer is answer A.