R(p)=-2p^2+4000p

Solution for R(p)=-2p^2+4000p equation:

(R)=-2R^2+4000R

We move all terms to the left:

(R)-(-2R^2+4000R)=0

We get rid of parentheses

2R^2-4000R+R=0

We add all the numbers together, and all the variables

2R^2-3999R=0

a = 2; b = -3999; c = 0;
Δ = b2-4ac
Δ = -39992-4·2·0
Δ = 15992001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15992001}=3999$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3999)-3999}{2*2}=\frac{0}{4}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3999)+3999}{2*2}=\frac{7998}{4}
=1999+1/2
$