20=4x^2

Solution for 20=4x^2 equation:

20=4x^2

We move all terms to the left:

20-(4x^2)=0

a = -4; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-4)·20
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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