Solve the Quadratic Equation: Finding X in 3x²-3x=6

Solve the given equation:

$3x^2-3x=6$

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

$3x^2-3x=6 \\ 3x^2-3x-6=0 \\$Note that all coefficients and the free term are multiples of 3, hence we'll divide both sides of the equation by 3:

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

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 those values:

$m\cdot n=-2\\ m+n=-1\\$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. Remember that the possible factors of 2 are 2 and 1, satisfying the second requirement mentioned. This along with the fact that the numbers we're looking for have different signs leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-2 \\ n=1 \end{cases}$

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

$x^2-x-2=0 \\ \downarrow\\ (x-2)(x+1)=0$

From here we'll remember that the product of expressions equals zero only if at least one of the multiplying expressions equals zero,

Therefore we obtain two simple equations which we solve by isolating the unknown in each:

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

or:

$x+1=0\\ \boxed{x=-1}$

Let's summarize the solution of the equation:

$3x^2-3x=6 \\ 3x^2-3x-6=0 \\ x^2-x-2=0 \\ \downarrow\\ (x-2)(x+1)=0 \\ \downarrow\\ x-2=0\rightarrow\boxed{x=2}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=2,-1}$

Therefore the correct answer is answer B.