Answer
The solutions are $-3$ and $4$.
Work Step by Step
$ 3x^{2}-3x-36=0\qquad$ ...factor out $3$ from left side.
$ 3(x^{2}-x-12)=0\qquad$ ...divide entire expression with $3$.
$ x^{2}-x-12=0\qquad$ ...rewrite the expression as $(kx+m)(lx+n)$ where $k$ and $l$ are two numbers
such that their sum is $-1$ and product is $12$.
$\qquad$ ...those numbers are $-4$ and $3$.
$\qquad$ ...$-4+3=-1,\ -4\cdot 3=-12$
$\qquad$ ...therefore,
$x^{2}-x-12=(x-4)(x+3)$
$\qquad$ ...apply the zero product property and solve for $x$.
$x-4=0$ or $x+3=0$
$x=4$ or $x=-3$
The solutions are $-3$ and $4$.