Answer
$x=\pm 2$ or $x=\pm 3i$
Work Step by Step
$ x^{4}+5x^{2}-36=0\qquad$...substitute $x^{2}$ for $u$
$ u^{2}+5u-36=0\qquad$... solve with the Quadractic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-5\pm\sqrt{25+144}}{2}$
$u=\displaystyle \frac{-5\pm\sqrt{169}}{2}$
$u=\displaystyle \frac{-5\pm 13}{2}$
$u=\displaystyle \frac{-5+13}{2}=\frac{8}{2}=4$ or $u=\displaystyle \frac{-5-13}{2}=\frac{-18}{2}=-9$
Bring back $x^{2}=u$.
$x^{2}=4$ or $x^{2}=-9$
$x=\pm 2$ or $x=\pm 3i$
