# How do you solve 3z^2 + z - 1 = 0 using the quadratic formula?

Apr 3, 2018

$x = \frac{- 1 \pm \sqrt{13}}{6}$

#### Explanation:

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

In your equation which is in the general form
$a {z}^{2} + b z + c = 0$

$a = 3 , b = 1 , c = - 1$

So putting it into the quadratic formula,

$x = \frac{- 1 \pm \sqrt{1 + 12}}{6}$
$x = \frac{- 1 \pm \sqrt{13}}{6}$