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