Given the roots,
2 and -1/3
Therefore we can write,
[tex](x-2)(x+\frac{1}{3})=0[/tex]On solving,
[tex]\begin{gathered} (x-2)(x+\frac{1}{3})=0 \\ x^2+\frac{1}{3}x-2x-\frac{2}{3}=0 \\ x^2+\frac{x-6x}{3}-\frac{2}{3}=0 \\ 3x^2-5x-2=0 \end{gathered}[/tex]