# How do you find the zeros, real and imaginary, of y=3x^2+31x+9 using the quadratic formula?

Jan 6, 2016

Identify an substitute the values of $a$, $b$ and $c$ into the quadratic formula to find:

$x = \frac{- 31 \pm \sqrt{853}}{6}$

#### Explanation:

$y = 3 {x}^{2} + 31 x + 9$ is in the form $a {x}^{2} + b x + c$ with $a = 3$, $b = 31$ and $c = 9$.

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 31 \pm \sqrt{{31}^{2} - \left(4 \cdot 3 \cdot 9\right)}}{2 \cdot 3}$

$= \frac{- 31 \pm \sqrt{961 - 108}}{6}$

$= \frac{- 31 \pm \sqrt{853}}{6}$