(x-4)(x-3)=x^2-7x+12

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Solution for (x-4)(x-3)=x^2-7x+12 equation: 



(x-4)(x-3)=x^2-7x+12

We move all terms to the left:

(x-4)(x-3)-(x^2-7x+12)=0

We get rid of parentheses

-x^2+(x-4)(x-3)+7x-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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