R(x)=80x+-0.4x2

Solution for R(x)=80x+-0.4x2 equation:

(R)=80R+-0.4R^2

We move all terms to the left:

(R)-(80R+-0.4R^2)=0

We use the square of the difference formula

-(80R-0.4R^2)+R=0

We get rid of parentheses

0.4R^2-80R+R=0

We add all the numbers together, and all the variables

0.4R^2-79R=0

a = 0.4; b = -79; c = 0;
Δ = b2-4ac
Δ = -792-4·0.4·0
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-79)-79}{2*0.4}=\frac{0}{0.8}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+79}{2*0.4}=\frac{158}{0.8}
=197+0.4/0.8
$