# How do you find the zeros, real and imaginary, of y=-2x^2-9x+5 using the quadratic formula?

Apr 22, 2018

Zeros are real and $x = 0.5 , x = - 5$

#### Explanation:

$y = - 2 {x}^{2} - 9 x + 5$

Comparing with standard quadratic equation $a {x}^{2} + b x + c = 0$

$a = - 2 , b = - 9 , c = 5$. Discriminant $D = {b}^{2} - 4 a c$ or

$D = 81 + 40 = 121$ , discriminant is positive, we get two real

solutions. Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{9 \pm \sqrt{121}}{- 4} = \frac{9 \pm 11}{-} 4 \therefore x = \frac{20}{-} 4 = - 5$ or

$x = \frac{- 2}{-} 4 = 0.5 \therefore$ Zeros are real and $x = 0.5 , x = - 5$ [Ans]