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3n^2+3n-4000=0
a = 3; b = 3; c = -4000;
Δ = b2-4ac
Δ = 32-4·3·(-4000)
Δ = 48009
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{48009}}{2*3}=\frac{-3-\sqrt{48009}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{48009}}{2*3}=\frac{-3+\sqrt{48009}}{6} $
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