(4b^2)-255=0

Solution for (4b^2)-255=0 equation:

(4b^2)-255=0

a = 4; b = 0; c = -255;
Δ = b2-4ac
Δ = 02-4·4·(-255)
Δ = 4080
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4080}=\sqrt{16*255}=\sqrt{16}*\sqrt{255}=4\sqrt{255}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{255}}{2*4}=\frac{0-4\sqrt{255}}{8}
=-\frac{4\sqrt{255}}{8}
=-\frac{\sqrt{255}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{255}}{2*4}=\frac{0+4\sqrt{255}}{8}
=\frac{4\sqrt{255}}{8}
=\frac{\sqrt{255}}{2}
$