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3n^2+3n-1002=0

a = 3; b = 3; c = -1002;

Δ = b^{2}-4ac

Δ = 3^{2}-4·3·(-1002)

Δ = 12033

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}$

The end solution:

$\sqrt{\Delta}=\sqrt{12033}=\sqrt{9*1337}=\sqrt{9}*\sqrt{1337}=3\sqrt{1337}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{1337}}{2*3}=\frac{-3-3\sqrt{1337}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{1337}}{2*3}=\frac{-3+3\sqrt{1337}}{6} $

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