Answer
The solution set is {$-2+5i, -2-5i$}.
Work Step by Step
$h(x) = 5(x+2)^2$, $h(x) = -125$
For $h(x) = 5(x+2)^2$, substitute $h(x)$ with $-125$.
$-125 = 5(x+2)^2$
Divide the equation by $5$. Thus, it becomes:
$-25 = (x+2)^2$
Use the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$.
$x+2= \sqrt{-25}$ or $x+2= -\sqrt{-25}$
$x= -2+\sqrt{-25}$ or $x= -2-\sqrt{-25}$
$x= -2+\sqrt{25}\sqrt {-1}$ or $x= -2-\sqrt{25}\sqrt {-1}$
$x = -2+5i$ or $x=-2-5i$.
The solution set is {$-2+5i, -2-5i$}.
