R=-4p^2+400p

Solution for R=-4p^2+400p equation:

=-4R^2+400R

We move all terms to the left:

-(-4R^2+400R)=0

We get rid of parentheses

4R^2-400R=0

a = 4; b = -400; c = 0;
Δ = b2-4ac
Δ = -4002-4·4·0
Δ = 160000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{160000}=400$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-400)-400}{2*4}=\frac{0}{8}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-400)+400}{2*4}=\frac{800}{8}
=100
$