Answer
$x=\displaystyle \pm\frac{\sqrt{5}}{2}$ or $x=\pm 1$
Work Step by Step
$ 4x^{4}-9x^{2}+5=0\qquad$...substitute $x^{2}$ for $u$
$ 4u^{2}-9u+5=0\qquad$... solve with the Quadractic formula. $a=4,\ b=-9,\ c=5$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-9,\ a$ for $4$ and $c$ for $5$.
$ u=\displaystyle \frac{-(-9)\pm\sqrt{(-9)^{2}-4\cdot(5)\cdot 4}}{2\cdot 4}\qquad$... simplify.
$u=\displaystyle \frac{9\pm\sqrt{81-80}}{8}$
$u=\displaystyle \frac{9\pm\sqrt{1}}{8}$
$ u=\displaystyle \frac{9\pm 1}{8}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{9+1}{8}=\frac{10}{8}=\frac{5}{4}$ or $u=\displaystyle \frac{9-1}{8}=\frac{8}{8}=1$
Bring back $x$.
$x^{2}=\displaystyle \frac{5}{4}$ or $x^{2}=1$
$x=\displaystyle \pm\frac{\sqrt{5}}{2}$ or $x=\pm 1$
