(9-z)(3z-2)=0

Solution for (9-z)(3z-2)=0 equation:

(9-z)(3z-2)=0

We add all the numbers together, and all the variables

(-1z+9)(3z-2)=0

We multiply parentheses ..

(-3z^2+2z+27z-18)=0

We get rid of parentheses

-3z^2+2z+27z-18=0

We add all the numbers together, and all the variables

-3z^2+29z-18=0

a = -3; b = 29; c = -18;
Δ = b2-4ac
Δ = 292-4·(-3)·(-18)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-25}{2*-3}=\frac{-54}{-6}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+25}{2*-3}=\frac{-4}{-6}
=2/3
$