(3z-4)(9-z)=0

Solution for (3z-4)(9-z)=0 equation:

(3z-4)(9-z)=0

We add all the numbers together, and all the variables

(3z-4)(-1z+9)=0

We multiply parentheses ..

(-3z^2+27z+4z-36)=0

We get rid of parentheses

-3z^2+27z+4z-36=0

We add all the numbers together, and all the variables

-3z^2+31z-36=0

a = -3; b = 31; c = -36;
Δ = b2-4ac
Δ = 312-4·(-3)·(-36)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-23}{2*-3}=\frac{-54}{-6}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+23}{2*-3}=\frac{-8}{-6}
=1+1/3
$