R(x)=0,1x^2+500x

Solution for R(x)=0,1x^2+500x equation:

(R)=0.1R^2+500R

We move all terms to the left:

(R)-(0.1R^2+500R)=0

We get rid of parentheses

-0.1R^2+R-500R=0

We add all the numbers together, and all the variables

-0.1R^2-499R=0

a = -0.1; b = -499; c = 0;
Δ = b2-4ac
Δ = -4992-4·(-0.1)·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*-0.1}=\frac{0}{-0.2}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-499)+499}{2*-0.1}=\frac{998}{-0.2}
=-4990
$