# How do you solve equation with quadratic formula 3n^2-4n-15=0?

Nov 1, 2016

n_1=-5/3; n_2=3

#### Explanation:

Since b-4 is even, you can use the formula:

$x = \frac{- \frac{b}{2} \pm \sqrt{{\left(\frac{b}{2}\right)}^{2} - a c}}{a}$, so you have

$n = \frac{- \left(- \frac{4}{2}\right) \pm \sqrt{{\left(\frac{4}{2}\right)}^{2} - 3 \cdot \left(- 15\right)}}{3}$
$= \frac{2 \pm \sqrt{4 + 45}}{3}$
$= \frac{2 \pm 7}{3}$
n_1=-5/3; n_2=3