r=-4r^2+48r+16

Solution for r=-4r^2+48r+16 equation:

r=-4r^2+48r+16

We move all terms to the left:

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

We get rid of parentheses

4r^2-48r+r-16=0

We add all the numbers together, and all the variables

4r^2-47r-16=0

a = 4; b = -47; c = -16;
Δ = b2-4ac
Δ = -472-4·4·(-16)
Δ = 2465
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{-(-47)-\sqrt{2465}}{2*4}=\frac{47-\sqrt{2465}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2465}}{2*4}=\frac{47+\sqrt{2465}}{8}
$