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(x-4)(x-3)2=x-7x+12
We move all terms to the left:
(x-4)(x-3)2-(x-7x+12)=0
We add all the numbers together, and all the variables
(x-4)(x-3)2-(-6x+12)=0
We get rid of parentheses
(x-4)(x-3)2+6x-12=0
We multiply parentheses ..
(+x^2-3x-4x+12)2+6x-12=0
We multiply parentheses
2x^2-6x-8x+6x+24-12=0
We add all the numbers together, and all the variables
2x^2-8x+12=0
a = 2; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·2·12
Δ = -32
Delta is less than zero, so there is no solution for the equation
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