Answer
$\{-2,-\sqrt 2,\sqrt 2,2\}$
Work Step by Step
We have to solve the quadratic equation:
$$x^4-6x^2+8=0.\tag1$$
We substitute $y=x^2$:
$$y^2-6y+8=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y-2)(y-4)&=0\\
y-2=0&\text{ or }y-4=0\\
y=2&\text{ or }y=4.
\end{align*}$$
We determine $x$ using the values of $y$ that we found:
$$\begin{align*}
x^2&=2\\
x&=\pm \sqrt 2\\
x^2&=4\\
x&=\pm2.
\end{align*}$$
The solution set of the equation is:
$$\{-2,-\sqrt 2,\sqrt 2,2\}.$$
