(4-z)(3z+5)=0

Solution for (4-z)(3z+5)=0 equation:

(4-z)(3z+5)=0

We add all the numbers together, and all the variables

(-1z+4)(3z+5)=0

We multiply parentheses ..

(-3z^2-5z+12z+20)=0

We get rid of parentheses

-3z^2-5z+12z+20=0

We add all the numbers together, and all the variables

-3z^2+7z+20=0

a = -3; b = 7; c = +20;
Δ = b2-4ac
Δ = 72-4·(-3)·20
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*-3}=\frac{-24}{-6}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*-3}=\frac{10}{-6}
=-1+2/3
$