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(x+2)(x-3)+(x+3)(x-2)=2(x-2)(x-3)
We move all terms to the left:
(x+2)(x-3)+(x+3)(x-2)-(2(x-2)(x-3))=0
We multiply parentheses ..
(+x^2-3x+2x-6)+(x+3)(x-2)-(2(x-2)(x-3))=0
We calculate terms in parentheses: -(2(x-2)(x-3)), so:We get rid of parentheses
2(x-2)(x-3)
We multiply parentheses ..
2(+x^2-3x-2x+6)
We multiply parentheses
2x^2-6x-4x+12
We add all the numbers together, and all the variables
2x^2-10x+12
Back to the equation:
-(2x^2-10x+12)
x^2-2x^2-3x+2x+(x+3)(x-2)+10x-6-12=0
We multiply parentheses ..
x^2-2x^2+(+x^2-2x+3x-6)-3x+2x+10x-6-12=0
We add all the numbers together, and all the variables
-1x^2+(+x^2-2x+3x-6)+9x-18=0
We get rid of parentheses
-1x^2+x^2-2x+3x+9x-6-18=0
We add all the numbers together, and all the variables
10x-24=0
We move all terms containing x to the left, all other terms to the right
10x=24
x=24/10
x=2+2/5
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