4z^2=-8z

Solution for 4z^2=-8z equation:

4z^2=-8z

We move all terms to the left:

4z^2-(-8z)=0

We get rid of parentheses

4z^2+8z=0

a = 4; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·4·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*4}=\frac{-16}{8}
=-2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*4}=\frac{0}{8}
=0
$