20z^2=4z

Solution for 20z^2=4z equation:

20z^2=4z

We move all terms to the left:

20z^2-(4z)=0

a = 20; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·20·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*20}=\frac{0}{40}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*20}=\frac{8}{40}
=1/5
$