#### Answer

The solution set is $\left\{-3, 3\right\}$.

#### Work Step by Step

Raise both sides of the equation to the 4th power to obtain:
$x^2+16 = (\sqrt{5})^4
\\x^2+16=25$
Subtract 16 on both sides of the equation to obtain:
$x^2=25-16
\\x^2=9$
Take the square root of both sides to obtain:
$x = \pm\sqrt{9}
\\x = \pm \sqrt{3^2}
\\x = \pm 3$
Thus, the solution set is $\left\{-3, 3\right\}$.