## Calculus with Applications (10th Edition)

$\displaystyle \frac{5}{2}$ and 1
$2r^{2}-7r+5=0$ Use the quadratic formula: (For $ax^{2}+bx+c=0,\ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ ) $r=\displaystyle \frac{-(-7)\pm\sqrt{(-7)^{2}-4(2)(5)}}{2(2)}$ $=\displaystyle \frac{7\pm\sqrt{49-40}}{4}=\frac{7\pm\sqrt{9}}{4}$ $r=\displaystyle \frac{7\pm 3}{4}$ $r=\displaystyle \frac{7+3}{4}$ or $r=\displaystyle \frac{7-3}{4}$ $r=\displaystyle \frac{10}{4}$ or $r=\displaystyle \frac{4}{4}$ $r=\displaystyle \frac{5}{2}$ or $r=1$ Solutions: $\displaystyle \frac{5}{2}$ and 1.