Answer
The solutions are $-2+\sqrt{10}$ and $-2-\sqrt{10}$.
Work Step by Step
$ 3x^{2}+12x-18=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $18$ to each side).
$ 3x^{2}+12x=18\qquad$ ...divide each term with $3$.
$ x^{2}+4x=6\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{4}{2})^{2}=2^{2}=4\qquad$ ...add $4$ to each side of the expression
$ x^{2}+4x+4=6+4\qquad$ ...simplify.
$ x^{2}+4x+4=10\qquad$ ... write left side as a binomial squared.
$(x+2)^{2}=10\qquad$ ...take square roots of each side.
$ x+2=\pm\sqrt{10}\qquad$ ...add $-2$ to each side
$ x+2-2=\pm\sqrt{10}-2\qquad$ ...simplify.
$x=-2\pm\sqrt{10}$
