The expression is given as,
[tex]2x^2\text{ < }3x\text{ + 2 }[/tex]
Simplifying the given expression ,
[tex]\begin{gathered} 2x^2-3x-2\text{ < 0} \\ (2x+1)(x-2)\text{ < }0 \\ (2x+1)\text{ < 0 or ( x - 2 ) < 0} \\ x\text{ > }\frac{-1}{2}\text{ or x < 2} \end{gathered}[/tex]
Thus the solution set is
[tex]\frac{-1}{2}\text{ < x < 2}[/tex]
