128=4k^2

Solution for 128=4k^2 equation:

128=4k^2

We move all terms to the left:

128-(4k^2)=0

a = -4; b = 0; c = +128;
Δ = b2-4ac
Δ = 02-4·(-4)·128
Δ = 2048
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*-4}=\frac{0-32\sqrt{2}}{-8}
=-\frac{32\sqrt{2}}{-8}
=-\frac{4\sqrt{2}}{-1}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*-4}=\frac{0+32\sqrt{2}}{-8}
=\frac{32\sqrt{2}}{-8}
=\frac{4\sqrt{2}}{-1}
$