Answer
$x=\pm 3$ or $x=\pm 2\sqrt{2}$
Work Step by Step
$(x^{2}-7)^{2}-3(x^{2}-7)+2=0\qquad$...substitute $x^{2}-7$ for $u$
$ u^{2}-3u+2=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{3\pm\sqrt{9-8}}{2}$
$u=\displaystyle \frac{3\pm\sqrt{1}}{2}$
$u=\displaystyle \frac{3\pm 1}{2}$
$u=\displaystyle \frac{3+1}{2}=\frac{4}{2}=2$ or $u=\displaystyle \frac{3-1}{2}=\frac{2}{2}=1$
Bring back $x^{2}-7=u$.
$x^{2}-7=2$ or $x^{2}-7=1$
$x^{2}=9$ or $x^{2}=8$
$x=\pm 3$ or $x=\pm 2\sqrt{2}$
