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=5Y^2-3Y-1
We move all terms to the left:
-(5Y^2-3Y-1)=0
We get rid of parentheses
-5Y^2+3Y+1=0
a = -5; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·(-5)·1
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{29}}{2*-5}=\frac{-3-\sqrt{29}}{-10} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*-5}=\frac{-3+\sqrt{29}}{-10} $
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