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

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Answer 1 · 2015-09-13T02:31:05

#x=(-9+sqrt33)/6, (-9-sqrt33)/6#

#(3x+3)(x+2)=2#

Expand the left side by using the foil method.

#3x^2+6x+3x+6=2#

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

Subtract #2# from both sides.

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

You now have a quadratic equation #ax^2+bx+c#, where #a=3, b=9, and c=4#.

Use the quadratic formula to solve for #x#.

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

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

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

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

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

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

#x=(-9+sqrt33)/6#

#x=(-9-sqrt33)/6#