z2=-18z-4z

Solution for z2=-18z-4z equation:

z2=-18z-4z

We move all terms to the left:

z2-(-18z-4z)=0

We add all the numbers together, and all the variables

z2-(-22z)=0

We add all the numbers together, and all the variables

z^2-(-22z)=0

We get rid of parentheses

z^2+22z=0

a = 1; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·1·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*1}=\frac{-44}{2}
=-22
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*1}=\frac{0}{2}
=0
$