Precalculus (6th Edition) Blitzer

Solution set = $\{-5, -3\}.$
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x^{2}+8x+15=0 \quad \rightarrow \left\{\begin{array}{l} a=1\\ b=8\\ c=15 \end{array}\right.$ $x=\displaystyle \frac{-8\pm\sqrt{8^{2}-4(1)(15)}}{2(1)}=\frac{-8\pm\sqrt{64-60}}{2}$ $=\displaystyle \frac{-8\pm\sqrt{4}}{2}=\frac{-8\pm 2}{2}$ $x=\displaystyle \frac{-8-2}{2}=\frac{-10}{2}=-5\qquad$ or $x=\displaystyle \frac{-8+2}{2}=\frac{-6}{2}=-3$ Solution set = $\{-5, -3\}.$