## College Algebra (10th Edition)

The solution set is $\left\{-7, 3\right\}$.
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 $\left\{-7, 3\right\}$.