Rewrite the equation as:
[tex]x^2-x+\frac{8}{6}=0[/tex]Let's use quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where:
a = 1
b = -1
c= 8/6
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(1)(\frac{8}{6})}}{2(1)} \\ \\ x=\frac{1\pm\sqrt[]{1-\frac{16}{3}}}{2} \\ \\ x=\frac{1\pm\sqrt[]{-\frac{13}{3}}}{2} \\ \\ x=\frac{1\pm\frac{\sqrt[]{39}}{3}i}{2} \\ \\ \end{gathered}[/tex]Simplify:
[tex]\begin{gathered} x=\frac{1}{6}(3-i\sqrt[]{39}) \\ or \\ x=\frac{1}{6}(3+i\sqrt[]{39}) \end{gathered}[/tex]