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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4z^2+33z-27=0

a = 4; b = 33; c = -27;
Δ = b2-4ac
Δ = 332-4·4·(-27)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-39}{2*4}=\frac{-72}{8}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+39}{2*4}=\frac{6}{8}
=3/4
$