Answer
$\left\{-7, 1\right\}$
Work Step by Step
RECALL:
To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation.
Thus, to solve the given equation by completing the square, add $(\frac{6}{2})^2=3^2=9$ on both sides of the equation to obtain:
$x^2+6x+9=7+9
\\x^2+6x+9=16
\\(x+3)^2=16$
Take the square root of both sides to obtain:
$x+3=\pm \sqrt{16}
\\x+3 = \pm\sqrt{4^2}
\\x+3=\pm4$
Subtract 3 on both sides of the equation to obtain:
$x = \pm4 - 3$
$x = 4-3 = 1$ or $x = -4-3 = -7$
Therefore, the solution of the given equation is $\left\{-7, 1\right\}$.
