Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula

Note that$$4+\frac1x-\frac1{x^2}=\frac1{x^2}\left(4x^2+x-1\right).$$So, solve the equation $4x^2+x-1=0$.

Multiplying both sides by $x^2$ will result in

$$4x^2+x-1=0$$

Now, with $a=4, b=1, c=-1$ use the quadratic formula and let us know what you get.

Other answers followed my suggestion in the comments.

Here's an alternative:

Let $z=\dfrac1x.$ Then we have $-z^2+z+4=0$, so, using the quadratic formula, $z=\dfrac{-1\pm\sqrt{17}}{-2}.$

Therefore $x=\dfrac1z=\dfrac{-2}{-1\pm\sqrt{17}}=\dfrac{2(-1\mp\sqrt{17})}{16}.$