n(3n+1)=400

Solution for n(3n+1)=400 equation:

n(3n+1)=400

We move all terms to the left:

n(3n+1)-(400)=0

We multiply parentheses

3n^2+n-400=0

a = 3; b = 1; c = -400;
Δ = b2-4ac
Δ = 12-4·3·(-400)
Δ = 4801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{4801}}{2*3}=\frac{-1-\sqrt{4801}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{4801}}{2*3}=\frac{-1+\sqrt{4801}}{6}
$