Is there a method for solving equations like $x^{2}-\sqrt{x}-2=0$? Is there a method for solving equations like $x^{2}-\sqrt{x}-2=0$?
As far as I can remember, I don't know any method for equations like this. 
 A: If you use the substitution $u=\sqrt x$ you see that your equation is basically a quartic. Since we can solve all polynomials of degree less than or equal to $4$, there is a "method" to solve it, but it's not necessarily easy.
As it turns out, there are two real solutions to $u^4-u-2=0$. The first is $-1$, but that turns out to be an extra root introduced when we did the substitution $u=\sqrt x$ (we cannot plug $-1$ back in to the formula or we have a $\sqrt {-1}$ term).
The other root is near $1.3$, If we solve this back for $x$, we get $x=1.3^2$ which is near $1.83$.
Thus, there is one real root to the original equation which is the square of the second root given by Wolfram here.
A: Setting $y = \sqrt{x}$, the equation becomes
$$
y^4 - y - 2 = 0 .
$$
This is a degree four polynomial equation. In general, it is possible to solve this kind of equations, but it is complicated and beyond the level a precalculus student should master; check the Wikipedia article on Quartic functions if you want to know more.
On a more basic level, it is useful to know that, once you spot an "easy" solution of the equation, finding the remaining solutions becomes easier because the order of the polynomial can be reduced. See the article on Synthetic division for a full explanation. In the case of the equation $y^4 - y -2 = 0$, it is easy to see that $-1$ is a solution. $y = -1$ does not correspond to a solution of your equation since it is negative, but that is not important. Using synthetic division, one obtains
$$
y^4 - y - 2 = (y + 1)(y^3 - y^2 + y - 2)
$$
so to find the other solutions to $y^4 - y -2 = 0$, it suffices to find the solutions of $y^3 - y^2 + y - 2 = 0$. This is slightly easier, since it is easier to find the roots of a cubic polynomial than to find those of a quartic polynomial, although it is still too hard for a precalculus course, I'm afraid.
A: A Mathematica code for $\displaystyle{x^{2} - \,\sqrt{\, x\, }\, - 2 = 0}$ which changes its sign in $\displaystyle{\left[1,2\right]}$ such that the 'starting guess' is $\displaystyle{{1 + 2 \over 2} = {3 \over 2}}$.

(* x^2 - Sqrt[x] - 2 = 0 *)

dRoot[x_] :=
  Module[{den, num, sqrtX},
   sqrtX = Sqrt[x];
   num = (2 - x^2) sqrtX + x;
   den = 4 x sqrtX - 1;
   2 num/den];

myTol = 1.05 Module[{mp, x = 1.0},
    Do[mp = x;
       x /= 2.0; 
       If[(1.0 + x) > 1.0, Continue[], Break[]], {100000}];
       Sqrt[mp]];

(* Newton-Rapson *)
 Module[{dr, r = 3/2, rAnt, tol},
  Do[rAnt = r;
     dr = dRoot[r];
     r += dr;
     tol = Abs[rAnt + dr/2];
     tol = If[tol > 0, Abs[dr]/tol, Abs[dr]];
     If[tol < myTol, Break[], Continue[]], {100000}
   ];
 N[r, 50]]

Root $\displaystyle{\approx \texttt{1.8311772072083418533865654461108296451696466030113}}$

$\displaystyle{\texttt{myTol} = 1.77017 \times 10^{-7}}$.


It agrees with $\texttt{@Claude Leibovici}$ answer.

