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

2 Answers
Jul 10, 2015

Answer:

#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 }#

Jul 10, 2015

Answer:

#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#