3n^2+n=310

Solution for 3n^2+n=310 equation:

3n^2+n=310

We move all terms to the left:

3n^2+n-(310)=0

a = 3; b = 1; c = -310;
Δ = b2-4ac
Δ = 12-4·3·(-310)
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-61}{2*3}=\frac{-62}{6}
=-10+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+61}{2*3}=\frac{60}{6}
=10
$