# How do you find the zeros, real and imaginary, of y=- 5x^2-22x+10  using the quadratic formula?

May 20, 2016

$\frac{11 \pm 3 \sqrt{19}}{5}$
$y = - 5 {x}^{2} - 22 x + 10 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 484 + 200 = 684 = 36 \left(19\right)$
$d = \pm 6 \sqrt{19}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{22}{10} \pm \frac{6 \sqrt{19}}{-} 10 = \frac{11 \pm 3 \sqrt{19}}{5}$