(2-3u)(2-u)=0

Solution for (2-3u)(2-u)=0 equation:

(2-3u)(2-u)=0

We add all the numbers together, and all the variables

(-3u+2)(-1u+2)=0

We multiply parentheses ..

(+3u^2-6u-2u+4)=0

We get rid of parentheses

3u^2-6u-2u+4=0

We add all the numbers together, and all the variables

3u^2-8u+4=0

a = 3; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·3·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*3}=\frac{4}{6}
=2/3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*3}=\frac{12}{6}
=2
$