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3a^2+a-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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*3}=\frac{-1-\sqrt{13}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*3}=\frac{-1+\sqrt{13}}{6} $
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