Answer
$-1$; $13 $
Work Step by Step
Given \begin{equation}
x^2-12 x-13=0.
\end{equation} Apply the method of completing the square to solve for $x$. Make sure that the coefficient of the square term is one before adding half the square of the coefficient of the $x$ term to both sides: \begin{equation}
\begin{aligned}
x^2-12 x-13 & =0 \\
x^2-12x&=13\\
x^2-2\cdot 6x+6^2 & =13+6^2 \\
(x-6)^2 & =49\\
x-6 & = \pm \sqrt{49} \\
x & =6 \pm 7.
\end{aligned}
\end{equation} This gives: \begin{equation}
\begin{aligned}
& x=6-7=-1 \\
& x=6+7=13.
\end{aligned}
\end{equation} Check: \begin{equation}
\begin{aligned}
(-1)^2-12(-1) -13& \stackrel{?}{=}0 \\
0 & =0\checkmark\\
(13)^2-12(13) -13& \stackrel{?}{=}0 \\
0 & =0\checkmark.
\end{aligned}
\end{equation} The solution is $$x=-1,\quad x=13.$$
