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8x(2x+2)=2x+(3x-2)
We move all terms to the left:
8x(2x+2)-(2x+(3x-2))=0
We multiply parentheses
16x^2+16x-(2x+(3x-2))=0
We calculate terms in parentheses: -(2x+(3x-2)), so:We get rid of parentheses
2x+(3x-2)
We get rid of parentheses
2x+3x-2
We add all the numbers together, and all the variables
5x-2
Back to the equation:
-(5x-2)
16x^2+16x-5x+2=0
We add all the numbers together, and all the variables
16x^2+11x+2=0
a = 16; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·16·2
Δ = -7
Delta is less than zero, so there is no solution for the equation
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