20=4z^2+1

Solution for 20=4z^2+1 equation:

20=4z^2+1

We move all terms to the left:

20-(4z^2+1)=0

We get rid of parentheses

-4z^2-1+20=0

We add all the numbers together, and all the variables

-4z^2+19=0

a = -4; b = 0; c = +19;
Δ = b2-4ac
Δ = 02-4·(-4)·19
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{19}}{2*-4}=\frac{0-4\sqrt{19}}{-8}
=-\frac{4\sqrt{19}}{-8}
=-\frac{\sqrt{19}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{19}}{2*-4}=\frac{0+4\sqrt{19}}{-8}
=\frac{4\sqrt{19}}{-8}
=\frac{\sqrt{19}}{-2}
$