## Calculus with Applications (10th Edition)

$\text{No real solutions.}$
$2x^{2}-7x+30=0$ Use the quadratic formula: (For $ax^{2}+bx+c=0,\ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ ) $x=\displaystyle \frac{-(-7)\pm\sqrt{(-7)^{2}-4(2)(30)}}{2(2)}$ $x=\displaystyle \frac{7\pm\sqrt{49-240}}{4}$ $x=\displaystyle \frac{7\pm\sqrt{-191}}{4}$ $\sqrt{-191}$ is not a real number. ( there is no real number whose square is $-191$) (squares of real numbers can not be negative) Therefore, there are no real solutions.