## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{1-\sqrt{29}}{4}, \displaystyle \frac{1+\sqrt{29}}{4}\}.$
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ Add $-2x-7$ to both sides, so we have $ax^{2}+bx+c=0.$ $4x^{2}-2x-7=0\quad\rightarrow \left\{\begin{array}{l} a=4\\ b=-2\\ c=-7 \end{array}\right.$ $x=\displaystyle \frac{2\pm\sqrt{(-2)^{2}-4(4)(-7)}}{2(4)}=\frac{2\pm\sqrt{4+112}}{8}=\frac{2\pm\sqrt{116}}{8}$ $=\displaystyle \frac{2\pm\sqrt{4\cdot 29}}{8}=\frac{2\pm 2\sqrt{29}}{8}=\frac{2(1\pm\sqrt{29})}{8}$ $=\displaystyle \frac{1\pm\sqrt{29}}{4}$ Solution set = $\displaystyle \{\frac{1-\sqrt{29}}{4}, \displaystyle \frac{1+\sqrt{29}}{4}\}.$