Example 12 Solve the equation
$$
5 x^{2}+x-x \sqrt{5 x^{2}-1}-2=0 \text {. }
$$
(1979, National High School Mathematics Competition)
Solution: Let $\sqrt{5 x^{2}-1}=y$, then the original equation can be transformed into $y^{2}-x y+x-1=0$.
Solving, we get $y=1$ or $y=x-1$.
Thus, $\sqrt{5 x^{2}-1}=1$ or $\sqrt{5 x^{2}-1}=x-1$.
$\therefore x= \pm \frac{\sqrt{10}}{5}$ or $x=-1$ or $x=\frac{1}{2}$.
Upon verification, $x= \pm \frac{\sqrt{10}}{5}$ are the roots of the original equation.
(3) Using constant substitution. Based on the characteristics of the equation, sometimes the constant can be treated as an "unknown", and the unknown can be treated as a "constant".