## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{-5-\sqrt{13}}{2},\frac{-5+\sqrt{13}}{2}\}.$
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x^{2}+5x+3=0 \quad\rightarrow \left\{\begin{array}{l} a=1\\ b=5\\ c=3 \end{array}\right.$ $x=\displaystyle \frac{-5\pm\sqrt{5^{2}-4(1)(3)}}{2(1)}=\frac{-5\pm\sqrt{25-12}}{2}=\frac{-5\pm\sqrt{13}}{2}$ $x= \displaystyle \frac{-5-\sqrt{13}}{2}\qquad$ or$\displaystyle \qquad x=\frac{-5+\sqrt{13}}{2}$ Solution set = $\displaystyle \{\frac{-5-\sqrt{13}}{2},\frac{-5+\sqrt{13}}{2}\}.$