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3n^2+3n-1000=0

a = 3; b = 3; c = -1000;

Δ = b^{2}-4ac

Δ = 3^{2}-4·3·(-1000)

Δ = 12009

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{-(3)-\sqrt{12009}}{2*3}=\frac{-3-\sqrt{12009}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{12009}}{2*3}=\frac{-3+\sqrt{12009}}{6} $

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