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

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

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

We add all the numbers together, and all the variables

(-1z+3)(3z-4)=0

We multiply parentheses ..

(-3z^2+4z+9z-12)=0

We get rid of parentheses

-3z^2+4z+9z-12=0

We add all the numbers together, and all the variables

-3z^2+13z-12=0

a = -3; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·(-3)·(-12)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-3}=\frac{-18}{-6}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-3}=\frac{-8}{-6}
=1+1/3
$