(x-4)(x-3)=x2-7x+12

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Solution for (x-4)(x-3)=x2-7x+12 equation:



(x-4)(x-3)=x2-7x+12
We move all terms to the left:
(x-4)(x-3)-(x2-7x+12)=0
We add all the numbers together, and all the variables
-(+x^2-7x+12)+(x-4)(x-3)=0
We get rid of parentheses
-x^2+7x+(x-4)(x-3)-12=0
We multiply parentheses ..
-x^2+(+x^2-3x-4x+12)+7x-12=0
We add all the numbers together, and all the variables
-1x^2+(+x^2-3x-4x+12)+7x-12=0
We get rid of parentheses
-1x^2+x^2-3x-4x+7x+12-12=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0

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