(4z+2)(8-z)=0

Solution for (4z+2)(8-z)=0 equation:

(4z+2)(8-z)=0

We add all the numbers together, and all the variables

(4z+2)(-1z+8)=0

We multiply parentheses ..

(-4z^2+32z-2z+16)=0

We get rid of parentheses

-4z^2+32z-2z+16=0

We add all the numbers together, and all the variables

-4z^2+30z+16=0

a = -4; b = 30; c = +16;
Δ = b2-4ac
Δ = 302-4·(-4)·16
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-34}{2*-4}=\frac{-64}{-8}
=+8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+34}{2*-4}=\frac{4}{-8}
=-1/2
$