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(3x+2)(x+1)=(x+4)(x-5)
We move all terms to the left:
(3x+2)(x+1)-((x+4)(x-5))=0
We multiply parentheses ..
(+3x^2+3x+2x+2)-((x+4)(x-5))=0
We calculate terms in parentheses: -((x+4)(x-5)), so:We get rid of parentheses
(x+4)(x-5)
We multiply parentheses ..
(+x^2-5x+4x-20)
We get rid of parentheses
x^2-5x+4x-20
We add all the numbers together, and all the variables
x^2-1x-20
Back to the equation:
-(x^2-1x-20)
3x^2-x^2+3x+2x+1x+2+20=0
We add all the numbers together, and all the variables
2x^2+6x+22=0
a = 2; b = 6; c = +22;
Δ = b2-4ac
Δ = 62-4·2·22
Δ = -140
Delta is less than zero, so there is no solution for the equation
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