# How do you solve 6x^2+7x+2=0?

Dec 11, 2016

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

#### Explanation:

You need to factor this quadratic by playing with multipliers for 6 (1x6, 2x3, 3x2, 6x1) and multipliers for 2 (1x2, 2x1):

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

Now we can solve each term for $0$:

$2 x + 1 = 0$

$2 x + 1 - 1 = 0 - 1$

$2 x = - 1$

$\frac{2 x}{2} = - \frac{1}{2}$

$x = - \frac{1}{2}$

and

$3 x + 2 = 0$

$3 x + 2 - 2 = 0 - 2$

$3 x = - 2$

$\frac{3 x}{3} = - \frac{2}{3}$

$x = - \frac{2}{3}$

Dec 11, 2016

$x = - \frac{1}{2} \text{ }$ or $\text{ } x = - \frac{2}{3}$

#### Explanation:

Given:

$6 {x}^{2} + 7 x + 2 = 0$

Use an AC method:

Find a pair of factors of $A C = 6 \cdot 2 = 12$ with sum $B = 7$.

The pair $4 , 3$ works.

Use this pair to split the middle term and factor by grouping:

$0 = 6 {x}^{2} + 7 x + 2$

$\textcolor{w h i t e}{0} = 6 {x}^{2} + 4 x + 3 x + 2$

$\textcolor{w h i t e}{0} = \left(6 {x}^{2} + 4 x\right) + \left(3 x + 2\right)$

$\textcolor{w h i t e}{0} = 2 x \left(3 x + 2\right) + 1 \left(3 x + 2\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 1\right) \left(3 x + 2\right)$

Hence zeros: $x = - \frac{1}{2}$ and $x = - \frac{2}{3}$