R(r)=12r-0.02r2

Solution for R(r)=12r-0.02r2 equation:

(R)=12R-0.02R^2

We move all terms to the left:

(R)-(12R-0.02R^2)=0

We get rid of parentheses

0.02R^2-12R+R=0

We add all the numbers together, and all the variables

0.02R^2-11R=0

a = 0.02; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·0.02·0
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*0.02}=\frac{0}{0.04}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*0.02}=\frac{22}{0.04}
=550
$