n(n+55)=2000

Solution for n(n+55)=2000 equation:

n(n+55)=2000

We move all terms to the left:

n(n+55)-(2000)=0

We multiply parentheses

n^2+55n-2000=0

a = 1; b = 55; c = -2000;
Δ = b2-4ac
Δ = 552-4·1·(-2000)
Δ = 11025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11025}=105$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-105}{2*1}=\frac{-160}{2}
=-80
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+105}{2*1}=\frac{50}{2}
=25
$