Answer
$x=\displaystyle \frac{1}{3}$ or $x=-\displaystyle \frac{1}{2}$
Work Step by Step
$ x^{-2}-x^{-1}-6=0\qquad$...substitute $x^{-1}$ for $u$ so that $u^{2}=x^{-2}$
$ u^{2}-u-6=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{1\pm\sqrt{1+24}}{2}$
$u=\displaystyle \frac{1\pm\sqrt{25}}{2}$
$u=\displaystyle \frac{1\pm 5}{2}$
$u=\displaystyle \frac{1+5}{2}=\frac{6}{2}=3$ or $u=\displaystyle \frac{1-5}{2}=\frac{-4}{2}=-2$
Bring back $x^{-1}=u$.
$x^{-1}=3$ or $ x^{-1}=-2\qquad$...recall that $x^{-1}=\displaystyle \frac{1}{x}$
$\displaystyle \frac{1}{x}=3$ or $\displaystyle \frac{1}{x}=-2\qquad$...multiply both sides by $x$.
$1=3x$ or $ 1=-2x\qquad$...divide the first expression with $3$ and the second with $-2$.
$x=\displaystyle \frac{1}{3}$ or $x=-\displaystyle \frac{1}{2}$
