How do you solve (3x+3)(x+2)=2 by factoring?

Sep 13, 2015

$x = \frac{- 9 + \sqrt{33}}{6} , \frac{- 9 - \sqrt{33}}{6}$

Explanation:

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

Expand the left side by using the foil method.

$3 {x}^{2} + 6 x + 3 x + 6 = 2$

$3 {x}^{2} + 9 x + 6 = 2$

Subtract $2$ from both sides.

$3 {x}^{2} + 9 x + 4 = 0$

You now have a quadratic equation $a {x}^{2} + b x + c$, where $a = 3 , b = 9 , \mathmr{and} c = 4$.

Use the quadratic formula to solve for $x$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the values for a, b, and c into the equation.

$x = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3} =$

$x = \frac{- 9 \pm \sqrt{81 \cdot - 48}}{6} =$

$x = \frac{- 9 \pm \sqrt{33}}{6}$

Separate the equation into two separate equations to find both solutions for $x$.

$x = \frac{- 9 + \sqrt{33}}{6}$

$x = \frac{- 9 - \sqrt{33}}{6}$