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(3x-2)(x+4)-x(x+1)=(2x+1)(x+4)-12
We move all terms to the left:
(3x-2)(x+4)-x(x+1)-((2x+1)(x+4)-12)=0
We multiply parentheses
-x^2+(3x-2)(x+4)-x-((2x+1)(x+4)-12)=0
We multiply parentheses ..
-x^2+(+3x^2+12x-2x-8)-x-((2x+1)(x+4)-12)=0
We calculate terms in parentheses: -((2x+1)(x+4)-12), so:We add all the numbers together, and all the variables
(2x+1)(x+4)-12
We multiply parentheses ..
(+2x^2+8x+x+4)-12
We get rid of parentheses
2x^2+8x+x+4-12
We add all the numbers together, and all the variables
2x^2+9x-8
Back to the equation:
-(2x^2+9x-8)
-1x^2+(+3x^2+12x-2x-8)-1x-(2x^2+9x-8)=0
We get rid of parentheses
-1x^2+3x^2-2x^2+12x-2x-1x-9x-8+8=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0
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