\[\boxed{\mathbf{12.}}\]
\[\sqrt{x^{2} + 6x - 2a} = x + 2\]
\[ОДЗ:\]
\[x + 2 \geq 0\]
\[x \geq - 2.\]
\[x^{2} + 6x - 2a = 0\]
\[D_{1} = 9 + 2a\]
\[x_{1} = - 3 + \sqrt{9 + 2a};\]
\[x_{2} = - 3 - \sqrt{9 + 2a};\]
\[x \geq - 2.\]
\[x^{2} + 6x - 2a = (x + 2)^{2}\]
\[x^{2} + 6x - 2a = x^{2} + 4x + 4\]
\[2x - 2a = 4\]
\[2x = 2a + 4\]
\[x = a + 2.\]
\[x \geq - 2:\]
\[a + 2 \geq - 2\]
\[a \geq - 4.\]
\[Ответ:при\ a \geq - 4.\]