Solve for x.

Solve for x.  Plese show all real and non-real solutions and show how you got to your answer.  If you end up with a radial in the numerator, rationalize the numerator.

$\begin{array}{|rcll|} \hline \frac{4\sqrt{3}}{3}{x}^{2}+\frac{4\sqrt{3}}{3}x &=& -\frac{4\sqrt{3}}{3} \qquad & | \qquad +\frac{4\sqrt{3}}{3}\\ \frac{4\sqrt{3}}{3}{x}^{2}+\frac{4\sqrt{3}}{3}x +\frac{4\sqrt{3}}{3} &=& 0 \qquad & | \qquad :\frac{4\sqrt{3}}{3}\\ x^2+ x + 1 &=& 0 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x_{1,2} &=& \frac{ b \pm \sqrt{b^2-4ac} } { 2a } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline x^2+ x + 1 &=& 0 \qquad a = 1 \qquad b=1 \qquad c = 1 \\ x_{1,2} &=& \frac{ 1 \pm \sqrt{1^2-4\cdot 1 \cdot 1} } { 2\cdot 1 } \\ x_{1,2} &=& \frac{ 1 \pm \sqrt{-3} } { 2 } \qquad & | \qquad \sqrt{-1} = i\\ x_{1,2} &=& \frac{ 1 \pm i \cdot \sqrt{3} } { 2 } \\\\ x_{1} &=& \frac{ 1 + i \cdot \sqrt{3} } { 2 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ \frac12 + i \cdot \frac{ \sqrt{3}}{2} }\\\\ x_{2} &=& \frac{ 1 - i \cdot \sqrt{3} } { 2 } \\ \mathbf{ x_{2} } &\mathbf{=} & \mathbf{ \frac12 - i \cdot \frac{ \sqrt{3}}{2} }\\ \hline \end{array}$