Answer
$x=1, x=-7$
Work Step by Step
We start by completing the square in line $1$ of the equation by dividing the coefficient of the $6x$ term by 2 and squaring it and adding the result on both sides.
$$
\begin{aligned}
x^2+6 x & = 7 \\
x^2+ 6x+\left(\frac{6}{2}\right)^2 & =7+\left(\frac{6}{2}\right)^2 \\
x^2+6x+3^2 & =7+3^2 \\
(x+3)^2 & =16.
\end{aligned}
$$ Take the square root:
$$
\begin{array}{rl}
(x+3)^2 & =16 \\
x+3 & = \pm \sqrt{16} \\
x+3 & = \pm 4 \\
x+3=4 &\ \textbf{or}\ \ \ x+3=-4\\
x=4-3 = 1& \ \textbf{or}\ \ \ x+3=-4-3= -7.
\end{array}
$$ Check $$\begin{aligned}
1^2+6 (1) & \stackrel{?}{=} 7\\
1+6&\stackrel{?}{=} 7\\
7& = 7\checkmark
\end{aligned}$$ $$\begin{aligned}
(-7)^2+6 (-7) & \stackrel{?}{=} 7\\
49-42& \stackrel{?}{=} 7\\
7& = 7\checkmark.
\end{aligned}$$ The solution is: $$x=1 \quad x=-7.$$
