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(4x+1)(x+3)-(x+5)(x-3)=(3x+1)(x-4)
We move all terms to the left:
(4x+1)(x+3)-(x+5)(x-3)-((3x+1)(x-4))=0
We multiply parentheses ..
(+4x^2+12x+x+3)-(x+5)(x-3)-((3x+1)(x-4))=0
We calculate terms in parentheses: -((3x+1)(x-4)), so:We get rid of parentheses
(3x+1)(x-4)
We multiply parentheses ..
(+3x^2-12x+x-4)
We get rid of parentheses
3x^2-12x+x-4
We add all the numbers together, and all the variables
3x^2-11x-4
Back to the equation:
-(3x^2-11x-4)
4x^2-3x^2+12x+x-(x+5)(x-3)+11x+3+4=0
We multiply parentheses ..
4x^2-3x^2-(+x^2-3x+5x-15)+12x+x+11x+3+4=0
We add all the numbers together, and all the variables
x^2-(+x^2-3x+5x-15)+24x+7=0
We get rid of parentheses
x^2-x^2+3x-5x+24x+15+7=0
We add all the numbers together, and all the variables
22x+22=0
We move all terms containing x to the left, all other terms to the right
22x=-22
x=-22/22
x=-1
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