sqrt(sqrt(x-5) + x) = 5

$$\\\sqrt{\sqrt{x-5} + x} = 5\\\\
\left[\sqrt{\sqrt{x-5} + x}\right]^2 = 5^2\\\\
\sqrt{x-5} + x = 25\\\\
\sqrt{x-5} = 25-x\\\\
(\sqrt{x-5})^2 = (25-x)^2\\\\
x-5= 625+x^2-50x \\\\
x^2-51x+630=0\\\\$$

$$\\\triangle=51^2-4*1*630=81\\\\
x=\frac{51\pm 9}{2}\\\\
x=\frac{51+ 9}{2}\qquad or\qquad x=\frac{51- 9}{2}\\\\
x=\frac{60}{2}\qquad or\qquad x=\frac{42}{2}\\\\
x=30 \qquad or\qquad x=21\\\\$$

Test

x=30      

$$\\\sqrt{\sqrt{30-5}+30}\\
=\sqrt{5+30}\\
=\sqrt{35}\\
\ne5$$        x=30 is not a solution

Test x=21

$$\\\sqrt{\sqrt{21-5}+21}\\
=\sqrt{4+21}\\
=\sqrt{25}\\
=5\qquad excellent$$

So  the answer is x=21

sqrt(sqrt(x-5) + x) = 5        x = 21

$$\sqrt{ \sqrt{x-5 } +x } = 5 \quad | \quad x = 21\\\\
\sqrt{ \sqrt{21-5 } +21 } = 5 \\\\
\sqrt{ \sqrt{16 } +21 } = 5 \\\\
\sqrt{ 4 +21 } = 5 \\\\
\sqrt{ 25 } = 5 \\\\
5 \stackrel{!}{=} 5 \\$$

Heureka did this partly by visual inspection - it made sense to do it that way.

I always look for the L O N G E S T solution. :)     LOL