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3y^2-2y-1=0
a = 3; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·3·(-1)
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*3}=\frac{-2}{6} =-1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*3}=\frac{6}{6} =1 $
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