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=6Y^2-Y-5
We move all terms to the left:
-(6Y^2-Y-5)=0
We get rid of parentheses
-6Y^2+Y+5=0
a = -6; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-6)·5
Δ = 121
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}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*-6}=\frac{-12}{-12} =1 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*-6}=\frac{10}{-12} =-5/6 $
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