# How do you solve 3x^2-10x+8=0?

Jan 21, 2017

The answer is $S = \left\{2 , \frac{4}{3}\right\}$

#### Explanation:

We compare this equation to

$a {x}^{2} + b x + c = 0$

$3 {x}^{2} - 10 x + 8 = 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = 100 - 4 \cdot 3 \cdot 8 = 100 - 96 = 4$

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

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

${x}_{1} = \frac{10 + 2}{6} = 2$

${x}_{2} = \frac{10 - 2}{6} = \frac{8}{6} = \frac{4}{3}$

Jan 21, 2017

$x = \frac{4}{3}$ and$x = 2$

#### Explanation:

$3 {x}^{2} - 10 x + 8 = 0$

$\left(3 x - 4\right) \left(x - 2\right)$

$3 x = 4$
multiply both sides by$\frac{1}{3}$

$x = \frac{4}{3}$

$x - 2 = 0$

$x = 2$
substitute $x = \frac{4}{3} \mathmr{and} x = 2$

$3 {\left(\frac{4}{3}\right)}^{2} - 10 \left(\frac{4}{3}\right) + 8 = 0$

$3 \times \frac{16}{9} - \frac{40}{3} + 8 = 0$

${\cancel{48}}^{16} / {\cancel{9}}^{3} - \frac{40}{3} + 8 = 0$

$5 \frac{1}{3} - 13 \frac{1}{3} + 8 = 0$

$- 8 + 8 = 0$

$3 {\left(2\right)}^{2} - 10 \left(2\right) + 8 = 0$

$12 - 20 + 8 = 0$

$- 8 + 8 = 0$