Answer
$m=\pm\sqrt{2}$ or $m=\pm 2i$
Work Step by Step
$(m^{2}+3)^{2}-4(m^{2}+3)-5=0\qquad$...substitute $m^{2}+3$ for $u$
$ u^{2}-4u-5=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{4\pm\sqrt{16+20}}{2}$
$u=\displaystyle \frac{4\pm\sqrt{36}}{2}$
$u=\displaystyle \frac{4\pm 6}{2}$
$u=\displaystyle \frac{4+6}{2}=\frac{10}{2}=5$ or $u=\displaystyle \frac{4-6}{2}=\frac{-2}{2}=-1$
Bring back $m^{2}+3=u$.
$m^{2}+3=5$ or $m^{2}+3=-1$
$m^{2}=2$ or $m^{2}=-4$
$m=\pm\sqrt{2}$ or $m=\pm 2i$
