R=-48f^2+3120f

Solution for R=-48f^2+3120f equation:

=-48R^2+3120R

We move all terms to the left:

-(-48R^2+3120R)=0

We get rid of parentheses

48R^2-3120R=0

a = 48; b = -3120; c = 0;
Δ = b2-4ac
Δ = -31202-4·48·0
Δ = 9734400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9734400}=3120$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3120)-3120}{2*48}=\frac{0}{96}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3120)+3120}{2*48}=\frac{6240}{96}
=65
$