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3y^2-y-1=0
We add all the numbers together, and all the variables
3y^2-1y-1=0
a = 3; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·3·(-1)
Δ = 13
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{-(-1)-\sqrt{13}}{2*3}=\frac{1-\sqrt{13}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{13}}{2*3}=\frac{1+\sqrt{13}}{6} $
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