Since $(3+2 \sqrt{2}) \cdot(3-2 \sqrt{2})=9-8=1$
$$
\Rightarrow(3-2 \sqrt{2})=\frac{1}{(3+2 \sqrt{2})}
$$
Now $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$
$$
\Rightarrow(3+2 \sqrt{2})^{x^2-4}+\frac{1}{(3+2 \sqrt{2})^{x^2-4}}=6
$$
Let $y=(3+2 \sqrt{2})^{x^2-4}$
$$
\Rightarrow y+\frac{1}{y}=6 \Rightarrow y^2-6 y+1=0
$$
$\Rightarrow y=3+2 \sqrt{2}$ or $y=3-2 \sqrt{2}$
$\Rightarrow(3+2 \sqrt{2})^{x^2-4}=(3+2 \sqrt{2})$ or
$$
(3+2 \sqrt{2})^{x^2-4}=\frac{1}{(3+2 \sqrt{2})}
$$
$\Rightarrow x^2-4=1$ or $x^2-4=-1$
$\Rightarrow x=\sqrt{5}$ or $x=3 i$
Since $x=3 i$ is imaginary
Hence $x=\sqrt{5}$
Therefore $x^4+x^2+5=35$