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

Solution for (2u-3)(2u-1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4u^2-8u+3=0

a = 4; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·4·3
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*4}=\frac{4}{8}
=1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8}
=1+1/2
$