Answer
$\displaystyle x=\frac{-1\pm\sqrt{17}}{8}$
Work Step by Step
$4+\displaystyle \frac{1}{x}-\frac{1}{x^{2}}=0$
We multiply by $x^2$:
$x^{2}(4+\frac{1}{x}-\frac{1}{x^{2}})=x^{2}*0$
$4x^{2}+x-1=0$
We solve using the quadratic formula ($a=4,\ b=1,\ c=-1$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-1\pm\sqrt{1^{2}-4*4*-1}}{2*4}$
$\displaystyle x=\frac{-1\pm\sqrt{1+16}}{8}$
$\displaystyle x=\frac{-1\pm\sqrt{17}}{8}$
