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

Let's solve the given equation:

$3x^2-3x=6$

$$First, let's organize the equation by moving terms 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, so we'll divide both sides of the equation by 3:

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

ow we notice 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:

$m\cdot n=-2\\ m+n=-1\\$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, and now we'll remember that the possible factors of 2 are 2 and 1, satisfying the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead 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'll get two simple equations and solve them 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.