How do you solve #3x(x+2)=2# using the quadratic formula?

1 Answer
Jul 3, 2015

Answer:

Convert the given equation to the form of a quadratic equation, #ax^2+bx+c=0#. Identify a, b, and c. Substitute the values into the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)#, and solve for x. There will be two solutions for x.

Explanation:

The equation needs to be in the form #ax^2+bx+c=0#.

#3x(x+2)=2#

#3x^2+6x=2#

#3x^2+6x-2=0#

Quadratic Formula

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

#a=3;##b=6;##c=-2#

#x=(-6+-sqrt(6^2-4*3*-2))/(2*3)#

#x=(-6+-sqrt(36+24))/(6)#

#x=(-6+-sqrt(60))/6#

#sqrt(60)=sqrt(4xx15)=2sqrt 15#

#x=(-6+-2sqrt15)/6#

#x=-1+(2sqrt 15)/3#

#x=-1-(2sqrt15)/3#