Answer
$2\pm 2i$.
Work Step by Step
${{x}^{2}}-4x+8=0$
Rewrite the expression ${{x}^{2}}-4x+8=0$ as,
${{x}^{2}}-4x=-8$
Now, add $4$ on both sides of the equation as,
$\begin{align}
& {{x}^{2}}-4x+4=-8+4 \\
& {{\left( x-2 \right)}^{2}}=-4
\end{align}$
Now taking the square root on both sides:
$\begin{align}
& \sqrt{{{\left( x-2 \right)}^{2}}}=\sqrt{-4} \\
& \left( x-2 \right)=\sqrt{-4}
\end{align}$
Now, from the principle of squares property,
$\left( x-2 \right)=2i$ or $\left( x-2 \right)=-2i$
Now, consider the term $\left( x-2 \right)=2i$ and solve as,
$\begin{align}
& \left( x-2 \right)=2i \\
& x=2+2i
\end{align}$
Now consider the term $\left( x-2 \right)=-2i$ and solve as,
$\begin{align}
& \left( x-2 \right)=-2i \\
& x=2-2i
\end{align}$
Thus, the value of $x$ from the provided expression ${{x}^{2}}-4x+8=0$ is $2\pm 2i$.
