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2n^2-13n-11=0
a = 2; b = -13; c = -11;
Δ = b2-4ac
Δ = -132-4·2·(-11)
Δ = 257
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{257}}{2*2}=\frac{13-\sqrt{257}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{257}}{2*2}=\frac{13+\sqrt{257}}{4} $
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