hello,

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

$\begin{array}{lcll} \sqrt{ \sqrt{x-5} + x } &=& 5 \qquad &|\qquad \text{(square both sides)} \\ \sqrt{x-5} + x &=& 5^2 \\ \sqrt{x-5} + x &=& 25 \qquad &|\qquad - x \\ \sqrt{x-5} &=& 25-x \\ \sqrt{x-5} &=& 25-x \qquad &|\qquad \text{(square both sides)} \\ x-5 &=& (25-x)^2 \\ x-5 &=& 25^2-2\cdot 25x+x^2 \\ x-5 &=&625-50x+x^2 \\ 625-50x+x^2 &=& x-5 \qquad &|\qquad - x+5 \\ 625-50x+x^2 - x+5 &=&0 \\ x^2 -51x +630 &=&0 \\ x_{1,2} &=& \frac12 \cdot ( 51\pm \sqrt{51^2-4\cdot 630} ) \\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{2601-2530} )\\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{81} )\\ x_{1,2} &=& \frac12 \cdot ( 51\pm 9 )\\\\ x_{1} &=& \frac12 \cdot ( 51 + 9 )\\ \mathbf{ x_{1} } &\mathbf{=}& \mathbf{30}\\\\ x_{2} &=& \frac12 \cdot ( 51 - 9 )\\ \mathbf{ x_{2} } &\mathbf{=}& \mathbf{21} \end{array}$

Solve for x:
sqrt(sqrt(x-5)+x) = 5

Raise both sides to the power of two:
sqrt(x-5)+x = 25

Subtract x from both sides:
sqrt(x-5) = 25-x

Raise both sides to the power of two:
x-5 = (25-x)^2

Expand out terms of the right hand side:
x-5 = x^2-50 x+625

Subtract x^2-50 x+625 from both sides:
-x^2+51 x-630 = 0

The left hand side factors into a product with three terms:
-((x-30) (x-21)) = 0

Multiply both sides by -1:
(x-30) (x-21) = 0

Split into two equations:
x-30 = 0 or x-21 = 0

Add 30 to both sides:
x = 30 or x-21 = 0

Add 21 to both sides:
x = 30 or x = 21

sqrt(sqrt(x-5)+x) => sqrt(sqrt(21-5)+21)  =  5:
So this solution is correct

sqrt(sqrt(x-5)+x) => sqrt(sqrt(30-5)+30)  =  sqrt(35) ~~ 5.91608:
So this solution is incorrect

The solution is:
Answer: | x = 21

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

The only real solution is 21

theyre amazing arnt they