(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)(z+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
$