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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4z^2+39z+27=0

a = 4; b = 39; c = +27;
Δ = b2-4ac
Δ = 392-4·4·27
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-33}{2*4}=\frac{-72}{8}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+33}{2*4}=\frac{-6}{8}
=-3/4
$