# How do you find the zeros, real and imaginary, of y=8x^2-4x-73 using the quadratic formula?

Mar 6, 2018

$x \approx 3.281 \mathmr{and} - 2.781$

#### Explanation:

$y = 8 {x}^{2} - 4 x - 73$

Let's first check the discriminant:
$\Delta = {\left(- 4\right)}^{2} - 4 \times 8 \times \left(- 73\right)$

$= 16 + 2336 = 2352$

Since $\Delta > 0$ $y$ will have 2 real roots.

$x = \frac{4 \pm \sqrt{2352}}{2 \times 8}$
$\approx \frac{4 \pm 48.4874}{16}$
$\approx 3.281 \mathmr{and} - 2.781$