16r^2-256=0

Solution for 16r^2-256=0 equation:

16r^2-256=0

a = 16; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·16·(-256)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128}{2*16}=\frac{-128}{32}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128}{2*16}=\frac{128}{32}
=4
$