\[y = \sqrt{6x - 2x^{2}} + \sqrt{8 - 5x}\]
\[1)\ 6x - 2x^{2} \geq 0\]
\[2x^{2} - 6x \leq 0\]
\[2x(x - 3) \leq 0.\]
\[2)\ 8 - 5x \geq 0\]
\[- 5x \geq - 8\]
\[5x \leq 8\]
\[x \leq 1,6.\]
\[Ответ:x \in \lbrack 0;1,6\rbrack.\]