27z^2+22z=0

Solution for 27z^2+22z=0 equation:

27z^2+22z=0

a = 27; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·27·0
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*27}=\frac{-44}{54}
=-22/27
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*27}=\frac{0}{54}
=0
$