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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4u^2+30u+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:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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