## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{3-\sqrt{65}}{4},\frac{3+\sqrt{65}}{4}\}$
Multiply the LHS, use FOIL $2x^{2}+2x-5x-5=2$ $2x^{2}-3x-7=0$ ... apply the quadratic formula, $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\quad \left\{\begin{array}{l} a=2\\ b=-3\\ c=-7 \end{array}\right.$ $x=\displaystyle \frac{3\pm\sqrt{(-3)^{2}-4(2)(-7)}}{2(2)}=\frac{3\pm\sqrt{9+56}}{4}=\frac{3\pm\sqrt{65}}{4}$ Solution set = $\displaystyle \{\frac{3-\sqrt{65}}{4},\frac{3+\sqrt{65}}{4}\}$