[tex]6x^2-x<2\implies6x^2-x-2<0\implies(3x-2)(2x+1)<0[/tex]
In order for the left side to be negative, we need either [tex]3x-2>0[/tex] and [tex]2x+1<0[/tex], or [tex]3x-2<0[/tex] and [tex]2x+1>0[/tex].
- If [tex]3x-2>0[/tex] and [tex]2x+1<0[/tex], then [tex]x>\frac23[/tex] and [tex]x<-\frac12[/tex]. There is no value of [tex]x[/tex] that meets both conditions, so this case offers no solutions.
- If [tex]3x-2<0[/tex] and [tex]2x+1>0[/tex], then [tex]x<\frac23[/tex] and [tex]x>-\frac12[/tex]. This means any number [tex]x[/tex] between -1/2 and 2/3 satisfies the inequality.
In interval notation, the solution set is
[tex]\boxed{-\dfrac12<x<\dfrac23}[/tex]
