Answer:
[tex]\large\boxed{z=4\ or\ z=16}[/tex]
Step-by-step explanation:
[tex]z-6\sqrt{z}+8=0\\\\\text{Domain}:\ z\geq0\\\\\text{substitute}\ \sqrt{z}=t\geq0,\ \text{then}\ z=t^2:\\\\t^2-6t+8=0\\\\t^2-4t-2t+8=0\\\\t(t-4)-2(t-4)=0\\\\(t-4)(t-2)=0\iff t-4=0\ \vee\ t-2=0\\\\t-4=0\to\sqrt{z}-4=0\qquad\text{add 4 to both sides}\\\\\sqrt{z}=4\iff\boxed{z=16}\\\\t-2=0\to\sqrt{z}-2=0\qquad\text{add 2 to both sides}\\\\\sqrt{z}=2\iff \boxed{z=4}[/tex]
