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(3x-4)=(4x^2-5x+6)
We move all terms to the left:
(3x-4)-((4x^2-5x+6))=0
We get rid of parentheses
3x-((4x^2-5x+6))-4=0
We calculate terms in parentheses: -((4x^2-5x+6)), so:We get rid of parentheses
(4x^2-5x+6)
We get rid of parentheses
4x^2-5x+6
Back to the equation:
-(4x^2-5x+6)
-4x^2+3x+5x-6-4=0
We add all the numbers together, and all the variables
-4x^2+8x-10=0
a = -4; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·(-4)·(-10)
Δ = -96
Delta is less than zero, so there is no solution for the equation
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