Question

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

Answer 1

`x = (-9 + sqrt(33)) / 6 or (-9 - sqrt(33)) / 6`

Explanation:

Simplify the right-hand-side by expanding,

`(3x+3)(x+2) = 3x^2 + 9x + 6`

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

`3x^2 + 9x + 4 = 0`

Solve this using the quadratic formula:

`x = (-9 +- sqrt(81 - 4*12))/6 = (-9 +- sqrt(33)) / 6`

Thus, `x = { (-9 + sqrt(33)) / 6, (-9 - sqrt(33)) / 6 }`

Answer 2

`x=(-9+sqrt(33))/6,` `(-9-sqrt(33))/6`

Explanation:

`(3x+3)(x+2)=2`

FOIL the left side of the equation.

`3x^2+6x+3x+6=2` =

`3x^2+9x+6=2`

Subtract `2` from both sides.

`3x^2+9x+6-2=0` =

`3x^2+9x+4=0`

The equation is now in the form of a quadratic equation: `ax^2+bx+c=0`, where `a=3;` `b=9;` and `c=4`.

Use the quadratic formula to solve this equation.

Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)` =

`x=(-9+-sqrt(9^2-4*3*4))/(2*3)` =

`x=(-9+-sqrt(81-48))/6` =

`x=(-9+-sqrt(33))/6`

Two answers for `x`.

`x=(-9+sqrt(33))/6`

`x=(-9-sqrt(33))/6`