1280000=c*(1000-2c)+1200000

Solution for 1280000=c*(1000-2c)+1200000 equation:

1280000=c(1000-2c)+1200000

We move all terms to the left:

1280000-(c(1000-2c)+1200000)=0

We add all the numbers together, and all the variables

-(c(-2c+1000)+1200000)+1280000=0


We calculate terms in parentheses: -(c(-2c+1000)+1200000), so:

c(-2c+1000)+1200000

We multiply parentheses

-2c^2+1000c+1200000

Back to the equation:

-(-2c^2+1000c+1200000)


We get rid of parentheses

2c^2-1000c-1200000+1280000=0

We add all the numbers together, and all the variables

2c^2-1000c+80000=0

a = 2; b = -1000; c = +80000;
Δ = b2-4ac
Δ = -10002-4·2·80000
Δ = 360000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{360000}=600$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1000)-600}{2*2}=\frac{400}{4}
=100
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1000)+600}{2*2}=\frac{1600}{4}
=400
$