2n(400-n)=1050

Solution for 2n(400-n)=1050 equation:

2n(400-n)=1050

We move all terms to the left:

2n(400-n)-(1050)=0

We add all the numbers together, and all the variables

2n(-1n+400)-1050=0

We multiply parentheses

-2n^2+800n-1050=0

a = -2; b = 800; c = -1050;
Δ = b2-4ac
Δ = 8002-4·(-2)·(-1050)
Δ = 631600
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:

$\sqrt{\Delta}=\sqrt{631600}=\sqrt{400*1579}=\sqrt{400}*\sqrt{1579}=20\sqrt{1579}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(800)-20\sqrt{1579}}{2*-2}=\frac{-800-20\sqrt{1579}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(800)+20\sqrt{1579}}{2*-2}=\frac{-800+20\sqrt{1579}}{-4}
$