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n^2+n-6480=0
a = 1; b = 1; c = -6480;
Δ = b2-4ac
Δ = 12-4·1·(-6480)
Δ = 25921
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25921}=161$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-161}{2*1}=\frac{-162}{2} =-81 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+161}{2*1}=\frac{160}{2} =80 $
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