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=3Y^2-3Y+5
We move all terms to the left:
-(3Y^2-3Y+5)=0
We get rid of parentheses
-3Y^2+3Y-5=0
a = -3; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·(-3)·(-5)
Δ = -51
Delta is less than zero, so there is no solution for the equation
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