11040*2=n(1200+40n-40)

Solution for 11040*2=n(1200+40n-40) equation:

11040*2=n(1200+40n-40)

We move all terms to the left:

11040*2-(n(1200+40n-40))=0

We add all the numbers together, and all the variables

-(n(40n+1160))+11040*2=0

We add all the numbers together, and all the variables

-(n(40n+1160))+22080=0


We calculate terms in parentheses: -(n(40n+1160)), so:

n(40n+1160)

We multiply parentheses

40n^2+1160n

Back to the equation:

-(40n^2+1160n)


We get rid of parentheses

-40n^2-1160n+22080=0

a = -40; b = -1160; c = +22080;
Δ = b2-4ac
Δ = -11602-4·(-40)·22080
Δ = 4878400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4878400}=\sqrt{1600*3049}=\sqrt{1600}*\sqrt{3049}=40\sqrt{3049}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1160)-40\sqrt{3049}}{2*-40}=\frac{1160-40\sqrt{3049}}{-80}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1160)+40\sqrt{3049}}{2*-40}=\frac{1160+40\sqrt{3049}}{-80}
$