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=6Y^2+3Y-11
We move all terms to the left:
-(6Y^2+3Y-11)=0
We get rid of parentheses
-6Y^2-3Y+11=0
a = -6; b = -3; c = +11;
Δ = b2-4ac
Δ = -32-4·(-6)·11
Δ = 273
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{273}}{2*-6}=\frac{3-\sqrt{273}}{-12} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{273}}{2*-6}=\frac{3+\sqrt{273}}{-12} $
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