(4-z)(2z+5)=0

Solution for (4-z)(2z+5)=0 equation:

(4-z)(2z+5)=0

We add all the numbers together, and all the variables

(-1z+4)(2z+5)=0

We multiply parentheses ..

(-2z^2-5z+8z+20)=0

We get rid of parentheses

-2z^2-5z+8z+20=0

We add all the numbers together, and all the variables

-2z^2+3z+20=0

a = -2; b = 3; c = +20;
Δ = b2-4ac
Δ = 32-4·(-2)·20
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{-(3)-13}{2*-2}=\frac{-16}{-4}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*-2}=\frac{10}{-4}
=-2+1/2
$