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=5Y^2+3Y-6
We move all terms to the left:
-(5Y^2+3Y-6)=0
We get rid of parentheses
-5Y^2-3Y+6=0
a = -5; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-5)·6
Δ = 129
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{129}}{2*-5}=\frac{3-\sqrt{129}}{-10} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*-5}=\frac{3+\sqrt{129}}{-10} $
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