Solution: We recall that sum of the roots of a quartic polynomial $a_4x^4 + a_3x^3 + a_2x^2 + a_1 x + a_0$ is $-\frac{a_3}{a_4}$. Let the roots of the give polynomial be $\alpha ,\beta ,\gamma ,\delta $. We have \[ \alpha + \beta + \gamma + \delta = -\frac{-3}{6} = \frac{1}{2}. \] Given that sum of two roots is zero. Let $\alpha + \beta = 0$ which implies $\alpha = - \beta $. Now we have \begin{equation}\label{eq:01Jul2023-1} \gamma + \delta = \frac{1}{2} \end{equation} We also have the following relations: \begin{align} & \alpha \beta + \beta \gamma + \gamma \delta + \delta \alpha = \frac{4}{3} \label{eq:01Jul2023-2} \\ & \alpha \beta \gamma + \beta \gamma \delta + \gamma \delta \alpha + \delta \alpha \beta = \frac{1}{6} \label{eq:01Jul2023-3} \\ & \alpha \beta \delta \gamma = \frac{1}{3} \label{eq:01Jul2023-4} \end{align}

From \eqref{eq:01Jul2023-1} and \eqref{eq:01Jul2023-3}, \begin{align*} \alpha \beta \gamma + \beta \gamma \delta + \gamma \delta \alpha + \delta \alpha \beta = \frac{1}{6} & \implies (\alpha +\beta)\gamma \delta + (\gamma +\delta) \alpha \beta = \frac{1}{6} \\ & \implies 0 + \frac{\alpha \beta }{2} = \frac{1}{6} \\ & \implies \alpha \beta = \frac{1}{3}. \end{align*} Substitute the value of $\alpha \beta $ in \eqref{eq:01Jul2023-4} to obtain \[ \gamma \delta = 1. \] Now we have $\beta =-\alpha $ and $\alpha \beta = \frac{1}{3}$ which implies \[ -\alpha ^2 = \frac{1}{3} \implies \alpha = \pm \iota \frac{1}{\sqrt{3} }. \] Similarly, $\gamma + \delta = \frac{1}{2}$ and $\gamma \delta =1$ implies \begin{align*} \gamma + \frac{1}{\gamma } = \frac{1}{2} & \implies 2\gamma ^2 - \gamma - 2 = 0 \\ & \implies \gamma = \frac{1 \pm \iota \sqrt{15} }{4}. \end{align*} Therefore, the roots are \[ \pm \frac{\iota }{\sqrt{3} }, \frac{1\pm \iota \sqrt{15} }{4}. \]