Question 587194
You're almost there.

I can see you are setting the denominator to 0 and you have:

{{{  x^2 + 10x + 25 = 0 }}}
Using Quadratic formula:  {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}   for the equation   {{{ ax^2 + bx + c }}}

We have a = 1 , b=10, c =25

   {{{x = (- 10 +- sqrt(10^2 - 4*1*25))/(2*1) }}}
or  {{{ x = (- 10 +- sqrt(100 - 100))/2 }}}
or  {{{ x = -10/2 }}}
or  x =-5

This shows that denominator becomes 0 for x= 5.

Therefore, we can say that x =-5 can not be in the domain of the given fucntion.

Hence, except for x =-5, all real numbers are the domain.
It is represented as:

x ϵ R - {-5}
or  x ϵ (-∞,∞) - {5}


Hope this helps~~