# How do you solve 3y^2 - 7y + 4 = 0 using the quadratic formula?

Jul 18, 2018

The solutions are $S = \left\{1 , \frac{4}{3}\right\}$

#### Explanation:

$3 {y}^{2} - 7 y + 4 = 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(- 7\right)}^{2} - 4 \cdot 3 \cdot 4 = 49 - 48 = 1$

As $\Delta > 0$, there are $2$ real roots

Therefore,

$y = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- \left(- 7\right) \pm \sqrt{1}}{2 \cdot 3}$

$= \frac{7 \pm 1}{6}$

Therefore,

${y}_{1} = \frac{8}{6} = \frac{4}{3}$

and

${y}_{2} = \frac{6}{6} = 1$

The solutions are $S = \left\{1 , \frac{4}{3}\right\}$

$y = \frac{4}{3} , 1$

#### Explanation:

$3 {y}^{2} - 7 y + 4 = 0$
$y = \setminus \frac{- \left(- 7\right) \setminus \pm \setminus \sqrt{{\left(- 7\right)}^{2} - 4 \left(3\right) \left(4\right)}}{2 \left(3\right)}$
$y = \setminus \frac{7 \setminus \pm 1}{6}$
$y = \frac{4}{3} , 1$