How do you solve #3(x^2-2x+5) -2(3x-5) = 4(x^2-x-5)-(x^2+3)#?

1 Answer
Oct 2, 2015

Answer:

#x=6#

Explanation:

First, expand each expression.
#3(x^2−2x+5)−2(3x−5)=4(x^2−x−5)−(x^2+3)#
#3x^2−6x+15−6x+10=4x^2−4x−20−x^2-3#

Next, combine everything in one side.
#3x^2−6x+15−6x+10=4x^2−4x−20−x^2-3#
#3x^2−6x+15−6x+10-4x^2+4x+20+x^2+3=0#

Combine all like terms.
#3x^2−6x+15−6x+10-4x^2+4x+20+x^2+3=0#
#3x^2-4x^2+x^2-6x-6x+4x+15+10+20+3=0#
#-8x+48=0#

Isolate #x#.
#-8x+48=0#
#-8x=-48#
#color(red)(x=6)#