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