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

Solution for (4u-1)(2-u)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4u^2+9u-2=0

a = -4; b = 9; c = -2;
Δ = b2-4ac
Δ = 92-4·(-4)·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*-4}=\frac{-16}{-8}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*-4}=\frac{-2}{-8}
=1/4
$