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=5+15Y-3Y^2
We move all terms to the left:
-(5+15Y-3Y^2)=0
We get rid of parentheses
3Y^2-15Y-5=0
a = 3; b = -15; c = -5;
Δ = b2-4ac
Δ = -152-4·3·(-5)
Δ = 285
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{-(-15)-\sqrt{285}}{2*3}=\frac{15-\sqrt{285}}{6} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{285}}{2*3}=\frac{15+\sqrt{285}}{6} $
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