R(p)=-0.4p^2+64p-2400

Solution for R(p)=-0.4p^2+64p-2400 equation:

(R)=-0.4R^2+64R-2400

We move all terms to the left:

(R)-(-0.4R^2+64R-2400)=0

We get rid of parentheses

0.4R^2-64R+R+2400=0

We add all the numbers together, and all the variables

0.4R^2-63R+2400=0

a = 0.4; b = -63; c = +2400;
Δ = b2-4ac
Δ = -632-4·0.4·2400
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-63)-\sqrt{129}}{2*0.4}=\frac{63-\sqrt{129}}{0.8}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+\sqrt{129}}{2*0.4}=\frac{63+\sqrt{129}}{0.8}
$