Answer:
[tex]x \leq 0\\or\\1 \leq x \leq \frac{3}{2}[/tex]
Step-by-step explanation:
[tex]2x^3-5x^2+3x \leq 0\\\Rightarrow x.(2x^2-5x+3) \leq 0\\\Rightarrow x.(2x^2-2x-3x+3) \leq 0\\\Rightarrow x.[2x(x-1)-3(x-1)] \leq 0\\\Rightarrow x(x-1)(2x-3) \leq 0\\Therefore\\ x \leq 0\\or\\1 \leq x \leq \frac{3}{2}[/tex]
