When you know the answer, it is often easier...
Indeed, begin with $(x-1)/\sqrt{x} = 2\sqrt{2}$
It is equivalent to $x-1 = 2\sqrt{2x}$
Equivalent to $(x-1)^2 = 8x$
Equivalent to $x^2-2x+1 = 8x$
Equivalent to $x^2-10x+1 = 0$
Equivalent to $(x-a)(x-b) = 0$ where $a = 5+2\sqrt{6}$ and $b = 5-2\sqrt{6}$
Equivalent to {$x=a$ or $x=b$}
Of course, since these are equivalences, this suffices to prove the fact. However, if you want to write it in a more straightforward way, it suffices to go back from the end to the beginning:
Let $x = 5+2\sqrt{6}$. Then $(x-(5+2\sqrt{6})(x-(5-2\sqrt{6}))=0$. But this product can be expanded as $x^2-10x+1$. So, we have $x^2+1 = 10x$. At this point, you can generate various identities. For instance, you could conclude $x^2+2x+1 = 12x$. Therefore $(x+1)^2=12x$ and $x+1 = 2\sqrt{3x}$ and so $(x+1)/\sqrt{x} = 2\sqrt{3}$...