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n^2+n-4000=0
a = 1; b = 1; c = -4000;
Δ = b2-4ac
Δ = 12-4·1·(-4000)
Δ = 16001
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{16001}}{2*1}=\frac{-1-\sqrt{16001}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{16001}}{2*1}=\frac{-1+\sqrt{16001}}{2} $
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