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