Answer
$ \lbrace0.58, 3.14\rbrace$
Work Step by Step
Let us consider the quadratic equation $x^{2}-4x=-2.$\\ To complete the square on the binomial $x^{2}-4x$, we take half of $-4$, which is $-2$ and square $-2$ giving $4$. We add $4$ to both sides of the equation. This makes the left side a perfect square binomial. Then
\begin{align*}
x^{2}-4x+4=-2+4.
\end{align*}
Rearranging left hand side, we get
\begin{align*}
x^{2}-2 \cdot x \cdot 2+2^{2}=2.
\end{align*}
The left side of above is in form of $a^{2}-2ab+b^{2}$,
\begin{align*}
(x-2)^{2}=2.
\end{align*}
According to square root property,
\begin{align*}
x-2&=\pm\sqrt{2}\\
x-2&=\sqrt{2} \ \ \text{or} \ \ x-2=-\sqrt{2}\\
x&=2+\sqrt{2}\ \ \text{or} \ \ x=2-\sqrt{2}.
\end{align*}
Hence, the solution are $3.41$ and $0.58$. The solution set is $ \lbrace0.58, 3.14\rbrace$
