# How do you find the zeros, real and imaginary, of y=-5x^2-3x+29 using the quadratic formula?

Jun 12, 2018

$\frac{- 3 \pm \sqrt{589}}{10}$

#### Explanation:

$y = - 5 {x}^{2} - 3 x + 29 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 9 + 580 = 589$ --> $d = \pm \sqrt{589}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{3}{-} 10 \pm \frac{\sqrt{589}}{10} = \frac{- 3 \pm \sqrt{589}}{10}$
$x 1 = \frac{- 3 + \sqrt{589}}{10}$
$x 2 = \frac{- 3 - \sqrt{589}}{10}$