R(p)=-2.5p^2+400p

Solution for R(p)=-2.5p^2+400p equation:

(R)=-2.5R^2+400R

We move all terms to the left:

(R)-(-2.5R^2+400R)=0

We get rid of parentheses

2.5R^2-400R+R=0

We add all the numbers together, and all the variables

2.5R^2-399R=0

a = 2.5; b = -399; c = 0;
Δ = b2-4ac
Δ = -3992-4·2.5·0
Δ = 159201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{159201}=399$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-399)-399}{2*2.5}=\frac{0}{5}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-399)+399}{2*2.5}=\frac{798}{5}
=159+3/5
$