7.215. $2^{x+\sqrt{x^{2}-4}}-5 \cdot(\sqrt{2})^{x-2+\sqrt{x^{2}-4}}-6=0$.
## Solution.

Domain of definition: $x^{2}-4 \geq 0 \Leftrightarrow x \in(-\infty ;-2] \cup[2 ; \infty)$.

Let's write the equation in the form $2^{x+\sqrt{x^{2}-4}}-\frac{5}{2} \cdot 2^{\frac{x+\sqrt{x^{2}-4}}{2}}-6=0$. Solving it as a quadratic equation in terms of $2^{\frac{x+\sqrt{x^{2}-4}}{2}}$, we get $2^{\frac{x+\sqrt{x^{2}-4}}{2}}=-\frac{3}{2}$ (no solutions), or $2^{\frac{x+\sqrt{x^{2}-4}}{2}}=2^{2} \Rightarrow \frac{x+\sqrt{x^{2}-4}}{2}=2, \sqrt{x^{2}-4}=4-x \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}x^{2}-4=16-8 x+x^{2}, \\ 4-x \geq 0,\end{array}\right.$

Answer: $\frac{5}{2}$.