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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2+2u+16u-8)=0

We get rid of parentheses

-4u^2+2u+16u-8=0

We add all the numbers together, and all the variables

-4u^2+18u-8=0

a = -4; b = 18; c = -8;
Δ = b2-4ac
Δ = 182-4·(-4)·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-14}{2*-4}=\frac{-32}{-8}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+14}{2*-4}=\frac{-4}{-8}
=1/2
$