\(3x^2-5x-2=0< =>3x^2-6x+x-2=0< =>3x\left(x-2\right)+\left(x-2\right)=0< =>\left(x-2\right)\left(3x+1\right)=0< =>\left[{}\begin{matrix}x-2=0\\3x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{3}\end{matrix}\right.\)
`3x^2-5x-2=0`
`=> 3x^2-6x+x-2=0`
`=> 3x(x-2)+(x-2)=0`
`=>(x-2)(3x+1)=0`
Trường hợp `1`
`x-2=0`
`=> x=0+2`
`=> x=2`
Trường hợp `2`
`3x+1=0`
`=> 3x=0-1`
`=> 3x=-1`
`=> x=(-1)/3`
`=> x=-1/3`
Vậy `x in {2;-1/3}`