Answer
$m=\displaystyle \frac{2}{3}$ or $m=-\displaystyle \frac{1}{5}$
Work Step by Step
$ 2m^{-2}+7m^{-1}-15=0\qquad$...substitute $m^{-1}$ for $u$ so that $u^{2}=m^{-2}$
$ 2u^{2}+7u-15=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-7\pm\sqrt{49+120}}{4}$
$u=\displaystyle \frac{-7\pm\sqrt{169}}{4}$
$u=\displaystyle \frac{-7\pm 13}{4}$
$u=\displaystyle \frac{-7+13}{4}=\frac{6}{4}=\frac{3}{2}$ or $u=\displaystyle \frac{-7-13}{4}=\frac{-20}{4}=-5$
Bring back $m^{-1}=u$.
$m^{-1}=\displaystyle \frac{3}{2}$ or $ m^{-1}=-5\qquad$...recall that $m^{-1}=\displaystyle \frac{1}{m}$
$\displaystyle \frac{1}{m}=\frac{3}{2}$ or $\displaystyle \frac{1}{m}=-5 $
$m=\displaystyle \frac{2}{3}$ or $m=-\displaystyle \frac{1}{5}$
