Answer
The solutions are $-6+3\sqrt{2}$ and $-6-3\sqrt{2}$.
Work Step by Step
$ x^{2}+12x+18=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $-18$ to each side).
$ x^{2}+12x+18-18=-18\qquad$ ...simplify.
$ x^{2}+12x=-18\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{12}{2})^{2}=6^{2}=36\qquad$ ...add $36$ to each side of the expression
$ x^{2}+12x+36=-18+36\qquad$ ...simplify.
$ x^{2}+12x+36=18\qquad$ ... write left side as a binomial squared.
$(x+6)^{2}=18\qquad$ ...take square roots of each side.
$ x+6=\pm\sqrt{18}\qquad$ ...rewrite $\sqrt{18}$ as $\sqrt{9\cdot 2}$
$ x+6=\pm\sqrt{9\cdot 2}\qquad$ ...evaluate $\sqrt{9}$.
$ x+6=\pm 3\sqrt{2}\qquad$ ...add $-6$ to each side.
$ x+6-6=\pm 3\sqrt{2}-6\qquad$ ...simplify.
$x=-6\pm 3\sqrt{2}$
