Answer
The solutions are $x=-8$ or $x=-4$.
Work Step by Step
$ x^{2}+12x+32=0\qquad$..add $-32$ to both sides so we can complete the square on the left side.
$ x^{2}+12x=-32\qquad$..add $36$ to both sides to complete the square ($\displaystyle \frac{1}{2}(12)=6$, and $(6)^{2}=36$.)
$ x^{2}+12x+36=-32+36\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(x+6)^{2}=4$
According to the general principle of square roots:
For any real number $k$ and any algebraic expression $x$ :
$\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$.
$ x+6=\pm\sqrt{4}\qquad$...add $-6$ to each side.
$ x+6-6=\pm\sqrt{4}-6\qquad$...simplify.
$x=-6\pm 2$
$x=-6+2$ or $t=-6-2$
$x=-4$ or $x=-8$
