R=-3p^2+60p+1060

Solution for R=-3p^2+60p+1060 equation:

=-3R^2+60R+1060

We move all terms to the left:

-(-3R^2+60R+1060)=0

We get rid of parentheses

3R^2-60R-1060=0

a = 3; b = -60; c = -1060;
Δ = b2-4ac
Δ = -602-4·3·(-1060)
Δ = 16320
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:

$\sqrt{\Delta}=\sqrt{16320}=\sqrt{64*255}=\sqrt{64}*\sqrt{255}=8\sqrt{255}$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-8\sqrt{255}}{2*3}=\frac{60-8\sqrt{255}}{6}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+8\sqrt{255}}{2*3}=\frac{60+8\sqrt{255}}{6}
$