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n^2+n-1640=0
a = 1; b = 1; c = -1640;
Δ = b2-4ac
Δ = 12-4·1·(-1640)
Δ = 6561
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{6561}=81$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-81}{2*1}=\frac{-82}{2} =-41 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+81}{2*1}=\frac{80}{2} =40 $
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