64r^2-16=0

Solution for 64r^2-16=0 equation:

64r^2-16=0

a = 64; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·64·(-16)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*64}=\frac{-64}{128}
=-1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*64}=\frac{64}{128}
=1/2
$