n(n+3)=108

Solution for n(n+3)=108 equation:

n(n+3)=108

We move all terms to the left:

n(n+3)-(108)=0

We multiply parentheses

n^2+3n-108=0

a = 1; b = 3; c = -108;
Δ = b2-4ac
Δ = 32-4·1·(-108)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*1}=\frac{-24}{2}
=-12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*1}=\frac{18}{2}
=9
$