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1-(2x+4)(x-3)=(x-5)(x-7)
We move all terms to the left:
1-(2x+4)(x-3)-((x-5)(x-7))=0
We multiply parentheses ..
-(+2x^2-6x+4x-12)-((x-5)(x-7))+1=0
We calculate terms in parentheses: -((x-5)(x-7)), so:We get rid of parentheses
(x-5)(x-7)
We multiply parentheses ..
(+x^2-7x-5x+35)
We get rid of parentheses
x^2-7x-5x+35
We add all the numbers together, and all the variables
x^2-12x+35
Back to the equation:
-(x^2-12x+35)
-2x^2-x^2+6x-4x+12x+12-35+1=0
We add all the numbers together, and all the variables
-3x^2+14x-22=0
a = -3; b = 14; c = -22;
Δ = b2-4ac
Δ = 142-4·(-3)·(-22)
Δ = -68
Delta is less than zero, so there is no solution for the equation
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