R(p)=-10p^2+600p

Solution for R(p)=-10p^2+600p equation:

(R)=-10R^2+600R

We move all terms to the left:

(R)-(-10R^2+600R)=0

We get rid of parentheses

10R^2-600R+R=0

We add all the numbers together, and all the variables

10R^2-599R=0

a = 10; b = -599; c = 0;
Δ = b2-4ac
Δ = -5992-4·10·0
Δ = 358801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{358801}=599$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-599)-599}{2*10}=\frac{0}{20}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-599)+599}{2*10}=\frac{1198}{20}
=59+9/10
$