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