Answer
The solutions are $1+2i\sqrt{6}$ and $1-2i\sqrt{6}$.
Work Step by Step
$ x^{2}-2x+25=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $-25$ to each side).
$ x^{2}-2x+25-25=0-25\qquad$ ...simplify.
$ x^{2}-2x=-25\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-2}{2})^{2}=(-1)^{2}=1\qquad$ ...add $1$ to each side of the expression
$ x^{2}-2x+1=-25+1\qquad$ ...simplify.
$ x^{2}-2x+1=-24\qquad$ ... write left side as a binomial squared.
$(x-1)^{2}=-24\qquad$ ...take square roots of each side.
$ x-1=\pm\sqrt{-24}\qquad$ ...simplify.($\sqrt{-24}=i\sqrt{24}$)
$ x-1=\pm i\sqrt{24}\qquad$ ...add $1$ to each side.
$ x-1+1=\pm i\sqrt{24}+1\qquad$ ...simplify.
$ x=1\pm i\sqrt{24}\qquad$ ...rewrite $\sqrt{24}$ as $\sqrt{4\cdot 6}$
$ x=1\pm i\sqrt{4\cdot 6.}\qquad$ ...evaluate $\sqrt{4}$.
$x=1\pm 2i\sqrt{6}$
