n(n-3)=550

Solution for n(n-3)=550 equation:

n(n-3)=550

We move all terms to the left:

n(n-3)-(550)=0

We multiply parentheses

n^2-3n-550=0

a = 1; b = -3; c = -550;
Δ = b2-4ac
Δ = -32-4·1·(-550)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-47}{2*1}=\frac{-44}{2}
=-22
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+47}{2*1}=\frac{50}{2}
=25
$