318. $\sqrt{x^{2}-1}=\sqrt{3}$.
Solution. Let's square both sides of the equation:

$$
\left(\sqrt{x^{2}-1}\right)^{2}=(\sqrt{3})^{2} ; x^{2}-1=3 ; x^{2}=4 ; x_{1}=2, x_{2}=-2
$$

We have obtained two solutions. Let's check each of them: if $x=2$, then $\sqrt{4-1}=\sqrt{3}$, i.e., $\sqrt{3}=\sqrt{3}$; if $x=-2$, then $\sqrt{4-1}=\sqrt{3}$, i.e., $\sqrt{3}=\sqrt{3}$. Thus, the values $x=2$ and $x_{2}=-2$ are roots of the given equation.