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20n^2-13n-2=0
a = 20; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·20·(-2)
Δ = 329
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{329}}{2*20}=\frac{13-\sqrt{329}}{40} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{329}}{2*20}=\frac{13+\sqrt{329}}{40} $
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