We can write the equation

\[x-8 \sqrt{x}+12=0 \]

by substituting \[y= \sqrt{x}\]

. The equation becomes \[y^2-8y+12=0\]

.This equation factorises as

\[(y-6)(y-2)=0\]

.Set each factor equal to 0 and solve.

\[y-6=0 \rightarrow y=6\]

\[y-2=0 \rightarrow y=2\]

\[y=6\]

then \[\sqrt{x}=6 \rightarrow x=6^2=36\]

.\[y=2\]

then \[\sqrt{x}=2 \rightarrow x=2^2=4\]

.