Answer
$x=\pm\sqrt{\frac{-5\pm\sqrt{37}}{6}}$
Work Step by Step
$ 3x^{4}+5x^{2}-1=0\qquad$...substitute $x^{2}$ for $u$
$ 3u^{2}+5u-1=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-5\pm\sqrt{25+12}}{6}$
$u=\displaystyle \frac{-5\pm\sqrt{37}}{6}$
$u=\displaystyle \frac{-5+\sqrt{37}}{6} $or $u=\displaystyle \frac{-5-\sqrt{37}}{6}$
Bring back $x^{2}=u$.
$x^{2}=\displaystyle \frac{-5+\sqrt{37}}{6}$ or $x^{2}=\displaystyle \frac{-5-\sqrt{37}}{6}$
$x=\pm\sqrt{\frac{-5\pm\sqrt{37}}{6}}$
