# How do you find the zeros, real and imaginary, of y=9x^2-4x-15 using the quadratic formula?

Jul 30, 2016

1.53 and - 1.09

#### Explanation:

$y = 9 {x}^{2} - 4 x - 15 = 0$
Use the new quadratic formula in graphic form (Socratic Search).
$D = {d}^{2} = {b}^{2} - 4 a c = 16 + 540 = 556$ --> $d = \pm 23.58$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{4}{18} \pm \frac{23.58}{18} = \frac{4 \pm 23.58}{18}$
$x 1 = \frac{27.58}{18} = 1.53$
$x 2 = - \frac{19.58}{18} = - 1. 09$
graph{9x^2 - 4x - 15 [-40, 40, -20, 20]}