r^2-48r+64=0

Solution for r^2-48r+64=0 equation:

r^2-48r+64=0

a = 1; b = -48; c = +64;
Δ = b2-4ac
Δ = -482-4·1·64
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32\sqrt{2}}{2*1}=\frac{48-32\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32\sqrt{2}}{2*1}=\frac{48+32\sqrt{2}}{2}
$