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

May 12, 2016

$x = - \frac{5}{3} \pm \frac{\sqrt{19}}{3}$

#### Explanation:

One way involves completing the square. To avoid fractions a little, premultiply by $3$ first, then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(3 x + 5\right)$ and $b = \sqrt{19}$ as follows:

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

$= 9 {x}^{2} + 30 x + 6$

$= {\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(5\right) + 6$

$= {\left(3 x + 5\right)}^{2} - 25 + 6$

$= {\left(3 x + 5\right)}^{2} - 19$

$= \left(3 x + 5\right) - {\left(\sqrt{19}\right)}^{2}$

$= \left(\left(3 x + 5\right) - \sqrt{19}\right) \left(\left(3 x + 5\right) + \sqrt{19}\right)$

$= \left(3 x + 5 - \sqrt{19}\right) \left(3 x + 5 + \sqrt{19}\right)$

$= 9 \left(x + \frac{5}{3} - \frac{\sqrt{19}}{3}\right) \left(x + \frac{5}{3} + \frac{\sqrt{19}}{3}\right)$

Hence:

$x = - \frac{5}{3} \pm \frac{\sqrt{19}}{3}$