$$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$, find $\lfloor z \rfloor.$
$z = -2 - \sqrt{3}$, so $\lfloor z \rfloor = \boxed{-4}$.
