Answer
$x=5-\sqrt{38}$ OR $x=5+\sqrt{38}$
Work Step by Step
Adding $\left( \dfrac{b}{2} \right)^2$ on both sides of the given equation, $
x^2-10x=13
,$ to complete the square of the left side results to
\begin{align*}
x^2-10x+\left( \dfrac{-10}{2} \right)^2&=13+\left( \dfrac{-10}{2} \right)^2
\\\\
x^2-10x+\left( -5 \right)^2&=13+\left( -5 \right)^2
\\\\
x^2-10x+25&=13+25
\\
(x-5)^2&=38
.\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-5&=\pm\sqrt{38}
\\
x&=5\pm\sqrt{38}
.\end{align*}
Hence, the solutions are $
x=5-\sqrt{38}$ OR $x=5+\sqrt{38}
$.
