Answer
$\text{No real solutions.}$
Work Step by Step
$2x^{2}-7x+30=0$
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.
