-4r2+48r+16=0

Solution for -4r2+48r+16=0 equation:

-4r^2+48r+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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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