Answer
The solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.
Work Step by Step
Subtract $5$ to both sides of the equation:
$x^2-2x+5-5=0-5
\\x^2-2x=-5$
Complete the square by adding $(\frac{-2}{2})^2=1$ to both sides of the equation:
$x^2-2x+1 = -5+1
\\x^2-2x+1=-4$
Write the trinomial in factored form to obtain:
$(x-1)^2=-4$
Take the square root of both sides:
$\sqrt{(x-1)^2} = \pm \sqrt{-4}
\\x-1=\pm \sqrt{-4}
\\x-1=\pm \sqrt{4(-1)}
\\x-1=\pm \sqrt{2^2(-1)}
\\x-1=\pm 2\sqrt{-1}$
Add $1$ to both sides of the equation to obtain:
$x-1+1=1\pm2\sqrt{-1}
\\x=1\pm 2\sqrt{-1}$
Note that $\sqrt{-1} = i$.
Thus, the equation above simplifies to:
$x=1 \pm 2i$
Therefore, the solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.
