Answer
$\left\{-4, -2\right\}$
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:
$x^2+6x+9=-8+9
\\x^2+6x+9=1
\\(x+3)^2=1$
Take the square root of both sides to obtain:
$x+3=\pm \sqrt{1}
\\x+3 = \pm1$
Subtract 3 on both sides of the equation to obtain:
$x = \pm1 - 3
\\x_1 = 1-3 = -2
\\x_2 = -1-3 = -4$
Therefore, the solution of the given equation is: $\left\{-4, -2\right\}$.
