6. Given that $\theta$ is a zero of $g(x)=x^{4}-7 x^{2}+1$, and $h(x)$ is a quadratic function with rational coefficients, satisfying $h\left(\theta^{2}-1\right)=0$. Then the roots of $h(x)=0$ are $\qquad$ .
6. $\frac{5 \pm 3 \sqrt{5}}{2}$.

Let $h(x)=a x^{2}+b x+c(a \neq 0)$.
Combining $\theta^{4}=7 \theta^{2}-1$ we get
$$
\begin{array}{l}
h\left(\theta^{2}-1\right)=0 \\
\Leftrightarrow a\left(\theta^{2}-1\right)^{2}+b\left(\theta^{2}-1\right)+c=0 \\
\Leftrightarrow a\left(\theta^{4}-2 \theta^{2}+1\right)+b\left(\theta^{2}-1\right)+c=0 \\
\Leftrightarrow(5 a+b) \theta^{2}-b+c=0 .
\end{array}
$$

By the problem, we know that $\theta^{2}$ is an irrational number.
$$
\text { Then }\left\{\begin{array}{c}
5 a+b=0, \\
-b+c=0
\end{array} \Leftrightarrow b=c=-5 a\right. \text {. }
$$

Therefore, $h(x)=a x^{2}-5 a x-5 a$
$$
=a\left(x^{2}-5 x-5\right) \text {. }
$$

Thus, the roots of $h(x)=0$ are $\frac{5 \pm 3 \sqrt{5}}{2}$.