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3y^2+9y-4=0
a = 3; b = 9; c = -4;
Δ = b2-4ac
Δ = 92-4·3·(-4)
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{-(9)-\sqrt{129}}{2*3}=\frac{-9-\sqrt{129}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*3}=\frac{-9+\sqrt{129}}{6} $
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