R(x)=x(12-0.04x)

Solution for R(x)=x(12-0.04x) equation:

(R)=R(12-0.04R)

We move all terms to the left:

(R)-(R(12-0.04R))=0

We add all the numbers together, and all the variables

R-(R(-0.04R+12))=0


We calculate terms in parentheses: -(R(-0.04R+12)), so:

R(-0.04R+12)

We multiply parentheses

0R^2+12R

We add all the numbers together, and all the variables

R^2+12R

Back to the equation:

-(R^2+12R)


We get rid of parentheses

-R^2+R-12R=0

We add all the numbers together, and all the variables

-1R^2-11R=0

a = -1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-1)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*-1}=\frac{0}{-2}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-1}=\frac{22}{-2}
=-11
$