Find the sums of the squares of the solutions of the equation 3z^2 - 5z + 8 = 0.

The roots of 3z^2 - 5z + 8 = 0 are $\frac{5 \pm i \sqrt{71}}{6}$

Then $(\frac{5 + i \sqrt{71}}{6})^2 + (\frac{5 - i \sqrt{71}}{6})^2 = -\frac{53}{36}$

Let the roots be x and y. Note that $x^2 + y^2 = ( x+ y)^2 - 2xy$

By Vieta's, $x + y = -{b \over a} = {5 \over 3}$ and $xy = {c \over a} = {8 \over 3}$

So, $x^2 + y^2 = {25 \over 9} - {16 \over 3} = {25 \over 9} - {48 \over 9} = \color{brown}\boxed{-{ 23 \over 9}}$