(9+z)(5z-4)=0

Solution for (9+z)(5z-4)=0 equation:

(9+z)(5z-4)=0

We add all the numbers together, and all the variables

(z+9)(5z-4)=0

We multiply parentheses ..

(+5z^2-4z+45z-36)=0

We get rid of parentheses

5z^2-4z+45z-36=0

We add all the numbers together, and all the variables

5z^2+41z-36=0

a = 5; b = 41; c = -36;
Δ = b2-4ac
Δ = 412-4·5·(-36)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-49}{2*5}=\frac{-90}{10}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+49}{2*5}=\frac{8}{10}
=4/5
$