Answer
$\left\{-\dfrac{1}{8},\dfrac{1}{7}\right\}$
Work Step by Step
We have to solve the quadratic equation:
$$x^{-2}+x^{-1}-56=0.\tag1$$
We substitute $y=x^{-1}$:
$$y^2+y-56=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y+8)(y-7)&=0\\
y+8=0&\text{ or }y-7=0\\
y=-8&\text{ or }y=7.
\end{align*}$$
We determine $x$ using the values of $y$ that we found.
$$\begin{align*}
y&=-8\Rightarrow x=y^{-1}=-\dfrac{1}{8}\\
y&=7\Rightarrow x=y^{-1}=\dfrac{1}{7}\\
\end{align*}$$
Check the solutions:
$$\begin{align*}
x&=-\dfrac{1}{8}\\
\left(-\dfrac{1}{8}\right)^{-2}+\left(-\dfrac{1}{8}\right)^{-1}-56&\stackrel{?}{=}0\\
64-8-56&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x&=\dfrac{1}{7}\\
\left(\dfrac{1}{7}\right)^{-2}+\left(\dfrac{1}{7}\right)^{-1}-56&\stackrel{?}{=}0\\
49+7-56&\stackrel{?}{=}0\\
0&=0\checkmark
\end{align*}$$
The solution set of the equation is:
$$\left\{-\dfrac{1}{8},\dfrac{1}{7}\right\}.$$
