(3z-4)(3z+4)=0

Solution for (3z-4)(3z+4)=0 equation:

(3z-4)(3z+4)=0

We use the square of the difference formula

9z^2-16=0

a = 9; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·9·(-16)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*9}=\frac{-24}{18}
=-1+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*9}=\frac{24}{18}
=1+1/3
$