(3z+1)(3z-1)=0

Solution for (3z+1)(3z-1)=0 equation:

(3z+1)(3z-1)=0

We use the square of the difference formula

9z^2-1=0

a = 9; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·9·(-1)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*9}=\frac{-6}{18}
=-1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*9}=\frac{6}{18}
=1/3
$