P=-4r2+48r+16

Solution for P=-4r2+48r+16 equation:

=-4P^2+48P+16

We move all terms to the left:

-(-4P^2+48P+16)=0

We get rid of parentheses

4P^2-48P-16=0

a = 4; b = -48; c = -16;
Δ = b2-4ac
Δ = -482-4·4·(-16)
Δ = 2560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{10}}{2*4}=\frac{48-16\sqrt{10}}{8}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{10}}{2*4}=\frac{48+16\sqrt{10}}{8}
$