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n^2+n-540=0
a = 1; b = 1; c = -540;
Δ = b2-4ac
Δ = 12-4·1·(-540)
Δ = 2161
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{-(1)-\sqrt{2161}}{2*1}=\frac{-1-\sqrt{2161}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2161}}{2*1}=\frac{-1+\sqrt{2161}}{2} $
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