$z=3x\left(3x-1\right)$

$x=\frac{\sqrt{4z+1}+1}{6}$
$x=\frac{-\sqrt{4z+1}+1}{6}$

$x=\frac{\sqrt{4z+1}+1}{6}$
$x=\frac{-\sqrt{4z+1}+1}{6}\text{, }z\geq -\frac{1}{4}$