2.304. $\sqrt{10 p+2 \sqrt{25 p^{2}-q^{2}}}-\sqrt{10 p-2 \sqrt{25 p^{2}-q^{2}}}=2 \sqrt{5 p-q}$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0127.jpg?height=324&width=780&top_left_y=664&top_left_x=169)

\[
\begin{aligned}
& \sqrt{10 p+2 \sqrt{25 p^{2}-q^{2}}}-\sqrt{10 p-2 \sqrt{25 p^{2}-q^{2}}}= \\
& =\sqrt{5 p+q+2 \sqrt{(5 p+q)(5 p-q)}+5 p-q}- \\
& -\sqrt{5 p+q-2 \sqrt{(5 p+q)(5 p-q)}+5 p-q}= \\
& =\sqrt{(\sqrt{5 p+q}+\sqrt{5 p-q})^{2}}-\sqrt{(\sqrt{5 p+q}-\sqrt{5 p-q})^{2}}= \\
& =\sqrt{5 p+q}+\sqrt{5 p-q}-|\sqrt{5 p+q}-\sqrt{5 p-q}|= \\
& =\left\{\begin{array}{l}
\sqrt{5 p+q}+\sqrt{5 p-q}+\sqrt{5 p+q}-\sqrt{5 p-q}=2 \sqrt{5 p+q} \text { for } q \geq 0 . \\
\sqrt{5 p+q}+\sqrt{5 p-q}-\sqrt{5 p+q}+\sqrt{5 p-q}=2 \sqrt{5 p-q} \text { for } q < 0 .
\end{array}\right.
\end{aligned}
\]

Answer: valid for all \( q \in [0 ; 5 p], p > 0 \).