3n+n^2=3304

Solution for 3n+n^2=3304 equation:

3n+n^2=3304

We move all terms to the left:

3n+n^2-(3304)=0

a = 1; b = 3; c = -3304;
Δ = b2-4ac
Δ = 32-4·1·(-3304)
Δ = 13225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{13225}=115$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-115}{2*1}=\frac{-118}{2}
=-59
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+115}{2*1}=\frac{112}{2}
=56
$