Answer
$x=1-i\sqrt{3}$ OR $x=1+i\sqrt{3}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
x^2-2x+4=0
,$ is equivalent to
\begin{align*}
x^2-2x+4-4=0-4
\\
x^2-2x=-4
.\end{align*}
Adding $\left( \dfrac{b}{2} \right)^2$ on both sides to complete the square of the left side results to
\begin{align*}
x^2-2x+\left(\dfrac{-2}{2} \right)^2&=-4+\left(\dfrac{-2}{2} \right)^2
\\\\
x^2-2x+\left( -1 \right)^2&=-4+\left(-1\right)^2
\\\\
x^2-2x+1&=-4+1
\\\\
(x-1)^2&=-3
.\end{align*}
Taking the square root of both sides (Square Root Property) and then solving for the variable, the equation above is equivalent to
\begin{align*}
x-1&=\pm\sqrt{-3}
\\\\
x-1&=\pm i\sqrt{3}
\\\\
x&=1\pm i\sqrt{3}
.\end{align*}
Hence, the solutions are
$x=1-i\sqrt{3}$ OR $x=1+i\sqrt{3}$.