Answer
$\left\{-6, 2\right\}$
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 $(\frac{4}{2})^2=2^2=4$ on both sides of the equation to obtain:
$x^2+4x+4=12+4
\\x^2+4x+4=16
\\(x+2)^2=16$
Take the square root of both sides to obtain:
$x+2=\pm \sqrt{16}
\\x+2=\pm \sqrt{4^2}
\\x+2=\pm 4$
Adad $-2$ on both sides of the equation to obtain:
$x = -2 \pm 4$
We get two solutions:
$x_1 = -2+4=2
\\x_2 = -2-4=-6$
Therefore, the solution of the given equation is: $\left\{-6, 2\right\}$.
