Answer
Solution set:$ \; \{-8, -6,4,6\}.$
Work Step by Step
By definition of the absolute value, $\; \left|A\right|=\left\{\begin{array}{lll}
A, & if & A\geq 0\\
-A & if & A\lt 0
\end{array}\right.$
This equation transforms into two equations without absolute values, $\left\{\begin{array}{llll}
(i) & x^{2}+2x-36=12 & for & x^{2}+2x-36\geq 0\\
(ii) & x^{2}+2x-36=-12 & for & x^{2}+2x-36\lt 0
\end{array}\right.$
$(i)$
$ x^{2}+2x-36=12$
$ x^{2}+2x-48=0$
$(x+8)(x-6)=0$
$ x=-8, \; x=6$
Test these solutions:
$\left[\begin{array}{lll}
|(-8)^{2}+2(-8)-36| & .... & |(6^{2}+2(6)-36| =\\
=|64-16-36| & & =|36+12-36|\\
=|12| & & =|12|\\
=12 & & =12
\end{array}\right]$
both are valid solutions.
$(ii)$
$ x^{2}+2x-36=-12$
$ x^{2}+2x-24=0$
$(x+6)(x-4)=0$
$ x=-6, \; x=4$
Test these solutions:
$\left[\begin{array}{lll}
|(-6)^{2}+2(-6)-36| & .... & |(4^{2}+2(4)-36| =\\
=|36-12-36| & & =|16+8-36|\\
=|-12| & & =|-12|\\
=12 & & =12
\end{array}\right]$
both are valid solutions.
Solution set:$ \; \{-8, -6,4,6\}.$
