Решите уравнение: корень из (2x+3)=6-x.

\[\sqrt{2x + 3} = 6 - x\]

\[6 - x \geq 0\]

\[x \leq 6.\]

\[\left( \sqrt{2x + 3} \right)^{2} = (6 - x)^{2}\]

\[2x + 3 = 36 - 12x + x^{2}\]

\[x^{2} - 14x + 33 = 0\]

\[D_{1} = 49 - 33 = 16\]

\[x_{1} = 7 + 4 = 11 > 6;\]

\[x_{2} = 7 - 4 = 3.\]

\[Ответ:x = 3.\]