Since the OP expressed some doubts about the procedure not the solution, here are some elements of reassurance. Let $a=10$, or, more generally, any positive real number.
In any interpretation of the exercise, the value of $x$, if it exists, should correspond to the limit of a sequence $(x_n)_{n\geqslant0}$ such that $x_{n+1}=\sqrt{a+\sqrt{a}x_n}$, for every $n\geqslant0$. Here is a fact:
Let $(x_n)_{n\geqslant0}$ denote any sequence defined as above. For every $x_0\geqslant-\sqrt{a}$, $x_n\to\ell_a$, where $\ell_a=\frac12(\sqrt5+1)\sqrt{a}$.
Thus, about any reasonable procedure used to define $x$, first, will succeed, and second, will yield $x=\ell_a$. We happy. (When $a=10$, $\ell_a=\frac{5+\sqrt5}{\sqrt{2}}$.)
To prove the fact stated above, consider the function $u_a$ defined by $u_a(t)=\sqrt{a+\sqrt{a}t}$, for every $t\geqslant-\sqrt{a}$. Then $u_a$ is continuous, increasing, such that $t\lt u_a(t)\lt\ell_a$ for every $-\sqrt{a}\leqslant t\lt\ell_a$, $u_a(\ell_a)=\ell_a$, and $\ell_a\lt u_a(t)\lt t$ for every $t\gt\ell_a$.
As a consequence, $x_n=\ell_a$ for every $n\geqslant0$ if $x_0=\ell_a$, $(x_n)_{n\geqslant0}$ is increasing and bounded above by $\ell_a$ and converges to $\ell_a$ for every $-\sqrt{a}\leqslant x_0\lt\ell_a$, and $(x_n)_{n\geqslant0}$ is decreasing and bounded below by $\ell_a$ and converges to $\ell_a$ for every $x_0\gt\ell_a$. In every case, $x_n\to\ell_a$.