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

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

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

We add all the numbers together, and all the variables

(-1z+4)(4z+5)=0

We multiply parentheses ..

(-4z^2-5z+16z+20)=0

We get rid of parentheses

-4z^2-5z+16z+20=0

We add all the numbers together, and all the variables

-4z^2+11z+20=0

a = -4; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·(-4)·20
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*-4}=\frac{-32}{-8}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*-4}=\frac{10}{-8}
=-1+1/4
$