$2.12 x^{2}+x+x^{-1}+x^{-2}=4$.

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$2.12 x^{2}+x+x^{-1}+x^{-2}=4$.
2.12 Rewriting the equation as

$$
\left(x^{2}+\frac{1}{x^{2}}\right)+\left(x+\frac{1}{x}\right)=4
$$

let $x+\frac{1}{x}=z$. Then $x^{2}+2+\frac{1}{x^{2}}=z^{2}$, or

$$
\begin{aligned}
& z^{2}-2=x^{2}+\frac{1}{x^{2}} . \text { The given equation becomes } \\
& z^{2}+z-6=0, \text { from which } z_{1}=2, z_{2}=-3
\end{aligned}
$$

Solution: $\quad x_{1}=1, x_{2,3}=\frac{-3 \pm \sqrt{5}}{2}$.