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

Solution for (1-z)(4z+9)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+1)(4z+9)=0

We multiply parentheses ..

(-4z^2-9z+4z+9)=0

We get rid of parentheses

-4z^2-9z+4z+9=0

We add all the numbers together, and all the variables

-4z^2-5z+9=0

a = -4; b = -5; c = +9;
Δ = b2-4ac
Δ = -52-4·(-4)·9
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*-4}=\frac{-8}{-8}
=1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*-4}=\frac{18}{-8}
=-2+1/4
$