R(x)=-1x^2+500x

Solution for R(x)=-1x^2+500x equation:

(R)=-1R^2+500R

We move all terms to the left:

(R)-(-1R^2+500R)=0

We get rid of parentheses

1R^2-500R+R=0

We add all the numbers together, and all the variables

R^2-499R=0

a = 1; b = -499; c = 0;
Δ = b2-4ac
Δ = -4992-4·1·0
Δ = 249001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{249001}=499$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-499)-499}{2*1}=\frac{0}{2}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-499)+499}{2*1}=\frac{998}{2}
=499
$