# How do you solve 15x^2=7x+2?

May 28, 2018

${x}_{1} = - \frac{1}{5}$ and ${x}_{2} = \frac{2}{3}$

#### Explanation:

$15 {x}^{2} = 7 x + 2$

$15 {x}^{2} - 7 x - 2 = 0$

$15 {x}^{2} - 10 x + 3 x - 2 = 0$

$5 x \cdot \left(3 x - 2\right) + 3 x - 2 = 0$

$\left(3 x - 2\right) \cdot \left(5 x + 1\right) = 0$

So ${x}_{1} = - \frac{1}{5}$ and ${x}_{2} = \frac{2}{3}$

May 28, 2018

${x}_{1} = \frac{2}{3}$ or ${x}_{2} = - \frac{1}{5}$
Dividing by $15$ we get
${x}^{2} - \frac{7}{15} x - \frac{2}{15} = 0$
${x}_{1 , 2} = \frac{7}{20} \pm \sqrt{\frac{49}{300} + \frac{2}{15}}$
${x}_{1 , 2} = \frac{7}{30} \pm \frac{13}{30}$
${x}_{1} = \frac{2}{3}$
${x}_{2} = - \frac{1}{5}$