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$\sqrt{3|x|+3}= \sqrt{x^2-25}\\ ( \sqrt{3|x|+3})^2= (\sqrt{x^2-25})^2\\3|x|+3=x^2-25\\x^2-3|x|-28=0\\\\1) x \leq 0\\x^2+3x-28=0\\x_1=-7\\x_2=4\; \; (\ \textgreater \ 0)\\x=-7\\\\2)x\ \textgreater \ 0\\x^2-3x-28=0\\x_1=7\\x_2=-4\; \; (\ \textless \ 0)\\x=7$

Проверка:
$1) x=-7\\ \sqrt{3|-7|+3}= \sqrt{(-7)^2-25}\\ \sqrt{3*7+3} = \sqrt{49-25}\\ \sqrt{24}= \sqrt{24}\; \; (verno)\\\\2)x=7\\\sqrt{3|7|+3}= \sqrt{7^2-25}\\ \sqrt{3*7+3} = \sqrt{49-25}\\ \sqrt{24}= \sqrt{24}\; \; (verno)$

Ответ: -7; 7