Find number of solutions of $|2x^2-5x+3|+x-1=0$ 
Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$
Solution:
Case 1: When $2x^2-5x+3 \geq 0$
Then we get, $2x^2-5x+3+x-1=0$
x=1,1
Case 2: When $2x^2-5x+3 < 0$
Then we get, $-2x^2+5x-3+x-1=0$
x=1,2
In both cases, common value of x is 1
Hence solution is x=1

Am I doing right ??
Somebody told me solution is x=2
 A: Nope, if you have $|f(x)| + g(x) = 0$ $(1)$ you solve it as follows:


*

*assume that $f(x)\geq 0$, find solutions of $f+g = 0$ - say $x_1$. Check whether $f(x_1)\geq 0$. If it holds, then $x_1$ solves $(1)$ as well. If it does not hold, it's not a solution of the original equation.

*assume $f(x)<0$, find solutions of $g-f = 0$ and check whether for them $f<0$.
In your case: for $f+g = 0$ you get a solution $x = 1$, which satisfies $f(1)\geq 0$, so that this is indeed one of the solutions of the original equation. For the case $f<0$ you get $x = 1$ or $x = 2$. We don't have to check $x = 1$ since we know it's already a solution, so we only check whether $f(2)<0$. Since the latter does not hold: $2\cdot 2^2 -5\cdot 2+3 = 1>0$, we obtain that $x=2$ is not a solution, and hence $1$ is the only solution.
A: $$|2x^2-5x+3|+x-1=
\begin{cases}
2x^2-5x+3+x-1 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
  -(2x^2-5x+3)+x-1 & 1\lt x\lt\frac32\\
\end{cases}\Rightarrow\begin{cases}
2x^2-4x+2 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
  -2x^2+6x-4 & 1\lt x\lt\frac32\\
\end{cases}$$

