#### Answer

$\color{blue}{\left\{-4\sqrt3, 4\sqrt3\right\}}$

#### Work Step by Step

Add $x^2$ to both sides to obtain:
$48=x^2
\\x^2=48$
RECALL:
If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$.
Take the square root of both sides of the given equation to obtain:
$\sqrt{x^2}=\pm \sqrt{48}
\\x =\pm \sqrt{16(3)}
\\x=\pm\sqrt{4^2(3)}
\\x=\pm 4\sqrt{3}$
Thus, the solution set is $\color{blue}{\left\{-4\sqrt3, 4\sqrt3\right\}}$.