(4z)^2=16z

Solution for (4z)^2=16z equation:

(4z)^2=16z

We move all terms to the left:

(4z)^2-(16z)=0

a = 4; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·4·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4}=\frac{0}{8}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4}=\frac{32}{8}
=4
$