Câu trả lời:

`(1/2+x)(2x-3)=0`

TH1: `1/2+x=0`

`=>x=0-1/2`

`=>x=-1/2`

TH2: `2x-3=0`

`=>2x=0+3`

`=>2x=3`

`=>x=3/2`

Vậy `x\in{-1/2;3/2}`