Answer
$m=\pm 1$ or $m=\pm 3i$
Work Step by Step
$(m^{2}+7)^{2}-6(m^{2}+7)-16=0\qquad$...substitute $m^{2}+7$ for $u$
$ u^{2}-6u-16=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{6\pm\sqrt{36+64}}{2}$
$u=\displaystyle \frac{6\pm\sqrt{100}}{2}$
$u=\displaystyle \frac{6\pm 10}{2}$
$u=\displaystyle \frac{6+10}{2}=\frac{16}{2}=8$ or $u=\displaystyle \frac{6-10}{2}=\frac{-4}{2}=-2$
Bring back $m^{2}+7=u$.
$m^{2}+7=8$ or $m^{2}+7=-2$
$m^{2}=1$ or $m^{2}=-9$
$m=\pm 1$ or $m=\pm 3i$
