R=-15p^2+300p+1200

Solution for R=-15p^2+300p+1200 equation:

=-15R^2+300R+1200

We move all terms to the left:

-(-15R^2+300R+1200)=0

We get rid of parentheses

15R^2-300R-1200=0

a = 15; b = -300; c = -1200;
Δ = b2-4ac
Δ = -3002-4·15·(-1200)
Δ = 162000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{162000}=\sqrt{32400*5}=\sqrt{32400}*\sqrt{5}=180\sqrt{5}$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-180\sqrt{5}}{2*15}=\frac{300-180\sqrt{5}}{30}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+180\sqrt{5}}{2*15}=\frac{300+180\sqrt{5}}{30}
$