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