Answer
$x=\pm 3$ or $x=\pm 2$
Work Step by Step
$ x^{4}-13x^{2}+36=0\qquad$...substitute $x^{2}$ for $u$
$ u^{2}-13u+36=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-13,\ c=36$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-13,\ a$ for $1$ and $c$ for $36$.
$ u=\displaystyle \frac{-(-13)\pm\sqrt{(-13)^{2}-4\cdot(36)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{13\pm\sqrt{169-144}}{2}$
$ u=\displaystyle \frac{13\pm\sqrt{25}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{13+5}{2}=\frac{18}{2}=9$ or $u=\displaystyle \frac{13-5}{2}=\frac{8}{2}=4$
Bring back $x$.
$x^{2}=9$ or $x^{2}=4$
$x=\pm 3$ or $x=\pm 2$
