# How do you solve 3y^2+5y-4=0 using the quadratic formula?

Apr 26, 2016

The solutions are $y = \frac{- 5 + \sqrt{73}}{6} , y = \frac{- 5 - \sqrt{73}}{6}$

#### Explanation:

$3 {y}^{2} + 5 y - 4 = 0$

The equation is of the form color(blue)(ay^2+by+c=0 where:

$a = 3 , b = 5 , c = - 4$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(5\right)}^{2} - \left(4 \cdot 3 \cdot \left(- 4\right)\right)$

$= 25 + 48 = 73$

The solutions are found using the formula
$y = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$y = \frac{\left(- 5\right) \pm \sqrt{73}}{2 \cdot 3} = \frac{- 5 \pm \sqrt{73}}{6}$

The solutions are:

• color(green)(y = (-5+sqrt(73))/6

• color(green)(y = (-5-sqrt(73))/6