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-2730+n+3n^2=0
a = 3; b = 1; c = -2730;
Δ = b2-4ac
Δ = 12-4·3·(-2730)
Δ = 32761
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{32761}=181$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-181}{2*3}=\frac{-182}{6} =-30+1/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+181}{2*3}=\frac{180}{6} =30 $
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