Answer
The solution set is $\left\{-7, 3\right\}$.
Work Step by Step
RECALL:
A quadratic equation in the form $x^2 + bx =c$ can be solved by completing the square by adding $\left(\frac{b}{2}\right)^2$ to both sides of the equation then using the square root property.
The equation will become $(x+\frac{b}{2})^2=c+(\frac{b}{2})^2$
Complete the square by adding $(\frac{4}{2})^2=4$ to both sides of the equation to obtain:
$x^2+4x+4=21+4
\\x^2+4x+4=25$
Write the trinomial as a square of a binomial to obtain:
$(x+2)^2=25$
Take the square root of both sides to obtain:
$x+2 = \pm\sqrt{25}
\\x+2 = \pm \sqrt{5^2}
\\x+2=\pm 5$
Add $-2$ to both sides to obtain:
$x+2+(-2) = -2 \pm 5
\\x = -2 \pm 5$
Split the solutions to obtain:
$x_1=-2+5
\\x_1=3$
$\\x_2=-2-5
\\x_2=-7$
Thus, the solution set is $\left\{-7, 3\right\}$.
