(4k)^2=36

Solution for (4k)^2=36 equation:

(4k)^2=36

We move all terms to the left:

(4k)^2-(36)=0

a = 4; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·4·(-36)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*4}=\frac{-24}{8}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*4}=\frac{24}{8}
=3
$