Answer
$x=\pm 2\sqrt{3}$ or $x=\pm 2$
Work Step by Step
$(x^{2}-2)^{2}-12(x^{2}-2)+20=0\qquad$...substitute $x^{2}-2$ for $u$
$ u^{2}-12u+20=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{12\pm\sqrt{144-80}}{2}$
$u=\displaystyle \frac{12\pm\sqrt{64}}{2}$
$u=\displaystyle \frac{12\pm 8}{2}$
$u=\displaystyle \frac{12+8}{2}=\frac{20}{2}=10$ or $u=\displaystyle \frac{12-8}{2}=\frac{4}{2}=2$
Bring back $x^{2}-2=u$.
$x^{2}-2=10$ or $x^{2}-2=2$
$x^{2}=12$ or $x^{2}=4$
$x=\pm 2\sqrt{3}$ or $x=\pm 2$
