Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 394: 110

{$-8,-6,4,6$}

Since, $|x^2+2x-36|=12$ Apply the definition of Absolute value. We have $x^2+2x-36=12$ and $x^2+2x-36=-12$ This implies that $x^2+2x-48=0$ and $x^2+2x-24=0$ $(x+8)(x-6)=0$ and $(x+6)(x-4)=0$ Thus, $x=-8,6$ and $x=-6,4$ Solution set is $x=${$-8,-6,4,6$}