16z^2=5

Solution for 16z^2=5 equation:

16z^2=5

We move all terms to the left:

16z^2-(5)=0

a = 16; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·16·(-5)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*16}=\frac{0-8\sqrt{5}}{32}
=-\frac{8\sqrt{5}}{32}
=-\frac{\sqrt{5}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*16}=\frac{0+8\sqrt{5}}{32}
=\frac{8\sqrt{5}}{32}
=\frac{\sqrt{5}}{4}
$