Question 709682
First we isolate the square root:
{{{sqrt(x+6)-4=x}}} --> {{{sqrt(x+6)=x+4}}}
Now, we square both sides of the equal sign,
knowing that we may be adding extraneous solutions,
but not worrying about them at this point.
{{{sqrt(x+6)=x+4}}} --> {{{x+6=(x+4)^2}}} --> {{{x+6=x^2+8x+16}}}
Now, we solve, still not worrying about extraneous solutions.
{{{x+6=x^2+8x+16}}} --> {{{0=x^2+7x+10}}}
If we are good at factoring, we factor
{{{x^2+7x+10=0}}} --> {{{(x+2)(x+5)=0}}}
and find the solutions to that equation as
{{{x=-2}}} and {{{x=-5}}}
(If we are not good at factoring,
we may solve the equation a different way,
such as "completing the square"
or using the quadratic formula).
 
At this point we worry about extraneous solutions,
but not to much.
We just check to see if the solutions we found for {{{x^2+7x+10=0}}}
are solutions to 
{{{sqrt(x+6)-4=x}}}
Substituting {{{x=-2}}}
{{{sqrt(x+6)-4=sqrt(-2+6)-4=sqrt(4)-4=2-4=-2}}}
so {{{highlight(x=-2)}}} is a solution of the original equation.
Substituting {{{x=-5}}}
{{{sqrt(x+6)-4=sqrt(-5+6)-4=sqrt(1)-4=1-4=-3}}}
so {{{highlight(x=-5)}}} is an extraneous solution.
It is not a solution of the original equation.