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9y^2+12y-11=0
a = 9; b = 12; c = -11;
Δ = b2-4ac
Δ = 122-4·9·(-11)
Δ = 540
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{15}}{2*9}=\frac{-12-6\sqrt{15}}{18} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{15}}{2*9}=\frac{-12+6\sqrt{15}}{18} $
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