R(x)=5x-0.3x^2

Solution for R(x)=5x-0.3x^2 equation:

(R)=5R-0.3R^2

We move all terms to the left:

(R)-(5R-0.3R^2)=0

We get rid of parentheses

0.3R^2-5R+R=0

We add all the numbers together, and all the variables

0.3R^2-4R=0

a = 0.3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·0.3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*0.3}=\frac{0}{0.6}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*0.3}=\frac{8}{0.6}
=13+0.2/0.6
$