(3z+5)(9-z)=0

Solution for (3z+5)(9-z)=0 equation:

(3z+5)(9-z)=0

We add all the numbers together, and all the variables

(3z+5)(-1z+9)=0

We multiply parentheses ..

(-3z^2+27z-5z+45)=0

We get rid of parentheses

-3z^2+27z-5z+45=0

We add all the numbers together, and all the variables

-3z^2+22z+45=0

a = -3; b = 22; c = +45;
Δ = b2-4ac
Δ = 222-4·(-3)·45
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-32}{2*-3}=\frac{-54}{-6}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+32}{2*-3}=\frac{10}{-6}
=-1+2/3
$