3n^2+n=2730

Solution for 3n^2+n=2730 equation:

3n^2+n=2730

We move all terms to the left:

3n^2+n-(2730)=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
$