n(n+3)=208

Solution for n(n+3)=208 equation:

n(n+3)=208

We move all terms to the left:

n(n+3)-(208)=0

We multiply parentheses

n^2+3n-208=0

a = 1; b = 3; c = -208;
Δ = b2-4ac
Δ = 32-4·1·(-208)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-29}{2*1}=\frac{-32}{2}
=-16
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+29}{2*1}=\frac{26}{2}
=13
$