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(2x^2=4x+1)+(3x^2+6x+5)
We move all terms to the left:
(2x^2-(4x+1)+(3x^2+6x+5))=0
We calculate terms in parentheses: +(2x^2-(4x+1)+(3x^2+6x+5)), so:We get rid of parentheses
2x^2-(4x+1)+(3x^2+6x+5)
We get rid of parentheses
2x^2+3x^2-4x+6x-1+5
We add all the numbers together, and all the variables
5x^2+2x+4
Back to the equation:
+(5x^2+2x+4)
5x^2+2x+4=0
a = 5; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·5·4
Δ = -76
Delta is less than zero, so there is no solution for the equation
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