Question 146271
I'm assuming you mean,
{{{H(x)=sqrt(25-x^2)}}}
The domain is every x where H(x) makes sense or is defined. 
The square root only makes sense or is defined when the argument,
that's the thing under the square root sign, is greater than or equal to zero.
{{{25-x^2>=0}}}
{{{25>=x^2}}}
Taking the square root of both sides, we get two answers. 
1.{{{x<=5}}}
2.{{{-x<=5}}}
From 2, 
{{{x>=-5}}}

So the domain of H(x) is 
{{{-5<=x<=5}}}