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

Jul 10, 2015

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

#### Explanation:

Simplify the right-hand-side by expanding,

$\left(3 x + 3\right) \left(x + 2\right) = 3 {x}^{2} + 9 x + 6$

Bring the 2 to the left hand side [Think of subtracting both sides by 2]. Then the equation becomes:

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

Solve this using the quadratic formula:

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 12}}{6} = \frac{- 9 \pm \sqrt{33}}{6}$

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

Jul 10, 2015

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

#### Explanation:

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

FOIL the left side of the equation.

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

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

Subtract $2$ from both sides.

$3 {x}^{2} + 9 x + 6 - 2 = 0$ =

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

The equation is now in the form of a quadratic equation: $a {x}^{2} + b x + c = 0$, where a=3; b=9; and $c = 4$.

Use the quadratic formula to solve this equation.

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

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

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

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

Two answers for $x$.

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

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