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6n^2+13n+1=0
a = 6; b = 13; c = +1;
Δ = b2-4ac
Δ = 132-4·6·1
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{145}}{2*6}=\frac{-13-\sqrt{145}}{12} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{145}}{2*6}=\frac{-13+\sqrt{145}}{12} $
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