(7z-9)(7z+9)=0

Solution for (7z-9)(7z+9)=0 equation:

(7z-9)(7z+9)=0

We use the square of the difference formula

49z^2-81=0

a = 49; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·49·(-81)
Δ = 15876
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15876}=126$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-126}{2*49}=\frac{-126}{98}
=-1+2/7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+126}{2*49}=\frac{126}{98}
=1+2/7
$