R(p)=-4p^2+1950p

Solution for R(p)=-4p^2+1950p equation:

(R)=-4R^2+1950R

We move all terms to the left:

(R)-(-4R^2+1950R)=0

We get rid of parentheses

4R^2-1950R+R=0

We add all the numbers together, and all the variables

4R^2-1949R=0

a = 4; b = -1949; c = 0;
Δ = b2-4ac
Δ = -19492-4·4·0
Δ = 3798601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3798601}=1949$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1949)-1949}{2*4}=\frac{0}{8}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1949)+1949}{2*4}=\frac{3898}{8}
=487+1/4
$