R(x)=-0.2x^2+20x

Solution for R(x)=-0.2x^2+20x equation:

(R)=-0.2R^2+20R

We move all terms to the left:

(R)-(-0.2R^2+20R)=0

We get rid of parentheses

0.2R^2-20R+R=0

We add all the numbers together, and all the variables

0.2R^2-19R=0

a = 0.2; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·0.2·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*0.2}=\frac{0}{0.4}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*0.2}=\frac{38}{0.4}
=95
$