3n+n^2=340

Solution for 3n+n^2=340 equation:

3n+n^2=340

We move all terms to the left:

3n+n^2-(340)=0

a = 1; b = 3; c = -340;
Δ = b2-4ac
Δ = 32-4·1·(-340)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-37}{2*1}=\frac{-40}{2}
=-20
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+37}{2*1}=\frac{34}{2}
=17
$