Many roots

Square the expression to get

$(25 + 10\sqrt{6}) + (25 - 10\sqrt{6}) + 2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})} =$

$50 + 2\sqrt{625 - 600} = 60$ .

So the given expression has a value of $\sqrt{60} = \boxed{2\sqrt{15}}.$

Let $a=\sqrt{25+10\sqrt{6} } , b= \sqrt{25-10 \sqrt{6}}$

$a^2 + b^2 = (25 + 10\sqrt6) + (25 -10\sqrt 6) = 50$ and $ab = \sqrt{(25+10\sqrt 6)(25- 10\sqrt 6)}=\sqrt {25^2 - (10\sqrt 6)^2}=\\ =\sqrt{625-600}=\sqrt{25}=5$

$(a+b)^2 = a^2 + b^2 + 2ab\\ =50+2\cdot 5\ =60\\ a+b=\sqrt{60}=2\sqrt{15}$

Let $a=25, b=10$ $\sqrt6$ , $x$ be the answer,

Hence,

$\sqrt{a+b}+\sqrt{a-b}=x$

Square both sides

$a+b+a-b+2\sqrt{a^2-b^2}=x^2$

$2(a+\sqrt{a^2-b^2}=x^2$

Substitute the values of a and b to get

$2(25+\sqrt{(25)^2-(10\sqrt6)^2}$ = $x^2$

$x^{2}=60$

$x=\sqrt{60}$

Therefore,

$\sqrt{25+10\sqrt6}$ + $\sqrt{25-10\sqrt6}$ = $\sqrt60$ = 2 $\sqrt{15}$