the number of solution to the equation |2x²-5x+3|+x-1=0 is?

2x² - 5x + 3 = (x - 1)(2x - 3)

Case I : When x ≤ 1 or x ≥ 3/2, then 2x² - 5x + 3 ≥ 0 and |2x² - 5x + 3| = 2x² - 5x + 3
|2x² - 5x + 3| + x - 1 = 0
2x² - 5x + 3 + x - 1 = 0
2x² - 4x + 2 = 0
x² - 2x + 1 = 0
(x - 1)² = 1
x = 1 (double roots)
In case I, x = 1

Case II : When 1 < x < 3/2, then 2x² - 5x + 3 < 0 and |2x² - 5x + 3| = -(2x² - 5x + 3)
|2x² - 5x + 3| + x - 1 = 0
-(2x² - 5x + 3) + x - 1 = 0
-2x² + 5x - 3 + x - 1 = 0
-2x² + 6x - 4 = 0
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 (rejected for 1 lies outside (1, 3/2)) or x = 2 (rejected) (rejected for 1 lies outside (1, 3/2))
In case II, there is no solution.

Combine Cases I and II. x = 1
There is only 1 solution.

　
|2x²−5x+3| + x − 1 = 0
|2x²−5x+3| = 1 − x

Since left side is ≥ 0, then so is right side:
1 − x ≥ 0
x ≤ 1

|2x²−5x+3| = 1 − x
2x² − 5x + 3 = 1 − x　　or　　2x² − 5x + 3 = −(1 − x)
2x² − 4x + 2 = 0　　or　　2x² − 6x + 4 = 0
2 (x² − 2x + 1) = 0　　or　　2 (x² − 3x + 2) = 0
2 (x − 1)² = 0　　or　　2 (x − 1) (x − 2) = 0
x = 2 ----> not a valid solution, since x must be ≤ 1
x = 1

One solution

1