Answer
The solution set is {$x=4+2i, x=4-2i$}.
Work Step by Step
$h(x) = 3(x-4)^2$, $h(x)= -12$
For $h(x) = 3(x-4)^2$, substitute $h(x)$ with $-12$.
$-12 = 3(x-4)^2$
Divide the equation by $3$. Thus, it becomes:
$-4 = (x-4)^2$
Use the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$.
$x-4 = \sqrt{-4}$ or $x-4 = -\sqrt{-4}$
$x = 4+\sqrt{-4}$ or $x = 4-\sqrt{-4}$
$x = 4+\sqrt4\sqrt{-1}$ or $x = 4-\sqrt4\sqrt{-1}$
$x=4+2i$ or $x=4-2i$
The solution set is {$x=4+2i, x=4-2i$}.
