# What are the zeros of the quadratic function f(x)=6x^2+12x-7?

Nov 19, 2016

$x = - 1 \pm \frac{1}{6} \sqrt{78}$

#### Explanation:

$f \left(x\right) = 6 {x}^{2} + 12 x - 7$

is of the form $a {x}^{2} + b x + c$ with $a = 6$, $b = 12$ and $c = - 7$

We can find the zeros of $f \left(x\right)$ using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \left(6\right) \left(- 7\right)}}{2 \left(6\right)}$

$\textcolor{w h i t e}{x} = \frac{- 12 \pm \sqrt{144 + 168}}{12}$

$\textcolor{w h i t e}{x} = \frac{- 12 \pm \sqrt{312}}{12}$

$\textcolor{w h i t e}{x} = \frac{- 12 \pm \sqrt{4 \cdot 78}}{12}$

$\textcolor{w h i t e}{x} = \frac{- 12 \pm 2 \sqrt{78}}{12}$

$\textcolor{w h i t e}{x} = - 1 \pm \frac{1}{6} \sqrt{78}$