# n^2+19n-500=0

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## Solution for n^2+19n-500=0 equation:

n^2+19n-500=0
a = 1; b = 19; c = -500;Δ = b2-4acΔ = 192-4·1·(-500)Δ = 2361The delta value is higher than zero, so the equation has two solutionsWe 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{-(19)-\sqrt{2361}}{2*1}=\frac{-19-\sqrt{2361}}{2}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{2361}}{2*1}=\frac{-19+\sqrt{2361}}{2}$

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