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=-1/3(Y-5)(Y+5)
We move all terms to the left:
-(-1/3(Y-5)(Y+5))=0
Domain of the equation: 3(Y-5)(Y+5))!=0We use the square of the difference formula
Y∈R
Y^2+25=0
a = 1; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·1·25
Δ = -100
Delta is less than zero, so there is no solution for the equation
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