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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3z^2+32z+45=0

a = 3; b = 32; c = +45;
Δ = b2-4ac
Δ = 322-4·3·45
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-22}{2*3}=\frac{-54}{6}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+22}{2*3}=\frac{-10}{6}
=-1+2/3
$