Question 981941
First isolate {{{sqrt(x)}}}



{{{x+sqrt(x)-6=0}}}



{{{x+sqrt(x)-6+6=0+6}}}



{{{x+sqrt(x)=6}}}



{{{sqrt(x)=6-x}}}



Now square both sides to get rid of the square root.



{{{sqrt(x)=6-x}}}



{{{(sqrt(x))^2=(6-x)^2}}}



{{{x=(6-x)^2}}}



{{{x=36-12x+x^2}}}



{{{x-x=36-12x+x^2-x}}}



{{{0=36-13x+x^2}}}



{{{x^2-13x+36=0}}}



{{{(x-9)(x-4)=0}}}



{{{x-9=0}}} or {{{x-4=0}}}



{{{x=9}}} or {{{x=4}}}



If you check both possible solutions back in the original equation, only {{{x = 4}}} works. The other value {{{x=9}}} does NOT satisfy the original equation. So it's not a true solution.