#### Answer

$\left\{-2, 2, -i, i\right\}$

#### Work Step by Step

Let
$u=x^2
\\u^2=x^4$
The given equation becomes:
$u^2-3u-4=0$
Factor the trinomial to obtain:
$(u-4)(u+1)=0$
Use the Zero-Factor Property by equating each factor to zero to obtain:
$u-4=0$ or $u+1=0$
Solve each equation:
$u=4$ or $u=-1$
Replace $u$ with $x^2$ to obtain:
$x^2=4$ or $x^2=-1$
Take the square root of both sides to obtain (note that $\sqrt{-1}=i$):
$x=\pm \sqrt{4}$ or $x=\pm \sqrt{-1}$
$x = \pm 2$ or $x = \pm i$
Therefore, the solution set is $\left\{-2, 2, -i, i\right\}$.