C=n(20-0.01n)+100

Solution for C=n(20-0.01n)+100 equation:

=C(20-0.01C)+100

We move all terms to the left:

-(C(20-0.01C)+100)=0

We add all the numbers together, and all the variables

-(C(-0.01C+20)+100)=0


We calculate terms in parentheses: -(C(-0.01C+20)+100), so:

C(-0.01C+20)+100

We multiply parentheses

0C^2+20C+100

We add all the numbers together, and all the variables

C^2+20C+100

Back to the equation:

-(C^2+20C+100)


We get rid of parentheses

-C^2-20C-100=0

We add all the numbers together, and all the variables

-1C^2-20C-100=0

a = -1; b = -20; c = -100;
Δ = b2-4ac
Δ = -202-4·(-1)·(-100)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$C=\frac{-b}{2a}=\frac{20}{-2}=-10$