theres is something i cant understamd

\(2x^2-3x+1\rightarrow(2x-1)(x-1)\)

This equals zero when x = 1/2 and x = 1  so 2x² - 3x + 1 is greater than zero when x is less than 1/2 and greater than 1.

2x²-3x+1>0      factor   and set to 0

(2x -1)(x - 1)  =  0       setting each factor to 0, we get that x = 1/2 and x = 1

Then, the solution(s)   lie  in one or more of these intervals :

( - infinity, 1/2), (1/2, 1) , (1, infinity )

Norrmally, if the middle interval solves the problem, the ourside two won't, or vice-versa

Let's test  point in the middle interval to see if it satisfies the original equation - I'll pick 3/4.....so we have

2(3/4)² -3(3/4) + 1 > 0

-1/8 > 0     and this is false.......so the other two intervals are the ones that solve the problem.......see the graph, here:  https://www.desmos.com/calculator/763e2oupae