Answer
$x=\pm\sqrt{\dfrac{-5+\sqrt{41}}{2}}$
Work Step by Step
$\sqrt[4]{4-5x^{2}}=x...$ Raise both sides to the 4th power
$4-5x^{2}=x^{4}$
$0=x^{4}+5x^{2}-4$
Substitute: $x^{2}=u$
$u^{2}+5u-4=0$
$u=\displaystyle \frac{-5\pm\sqrt{25+16}}{2}=\frac{-5\pm\sqrt{41}}{2}$
$u_{1}=\displaystyle \frac{-5-\sqrt{41}}{2}, u_{2}$=$\displaystyle \frac{-5+\sqrt{41}}{2}$
$x^{2}=\displaystyle \frac{-5-\sqrt{41}}{2}$ has no solutions
$x^{2}=\displaystyle \frac{-5+\sqrt{41}}{2}$
$x=\pm\sqrt{\dfrac{-5+\sqrt{41}}{2}}$