2. Given $x=\frac{-\sqrt[1]{17}+\sqrt{\sqrt{17}+4 \sqrt{15}}}{2 \sqrt{3}}$. Then, $3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5=$
2. 0 .

From the given, we have
$$
2 \sqrt{3} x+\sqrt[4]{17}=\sqrt{\sqrt{17}+4 \sqrt{15}} \text {. }
$$

Squaring and rearranging, we get
$$
\begin{array}{l}
\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}=0 . \\
\text { Also, } 3 x^{4}-(2 \sqrt{15}+\sqrt{17}) x^{2}+5 \\
=\left(3 x^{4}-2 \sqrt{3} \cdot \sqrt{5} x^{2}-5\right)-1-\sqrt{17} x^{2} \\
=\left(\sqrt{3} x^{2}-\sqrt{5}\right)^{2}-(\sqrt[4]{17} x)^{2} \\
=\left(\sqrt{3} x^{2}+\sqrt[4]{17} x-\sqrt{5}\right)\left(\sqrt{3} x^{2}-\right. \\
\sqrt[4]{17} x-\sqrt{5}) \\
\stackrel{1}{=} 0 .
\end{array}
$$

Explanation: Generally, if $x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$, then
$$
\begin{array}{l}
a^{2} x^{4}+\left(2 a c-b^{2}\right) x^{2}+c^{2} \\
=\left(a x^{2}+b x+c\right)\left(a x^{2}-b x+c\right) \\
=0 .
\end{array}
$$