\[\sqrt{44 - x} = x - 2\]
\[x - 2 \geq 0\]
\[x \geq 2.\]
\[44 - x = (x - 2)^{2}\]
\[44 - x = x^{2} - 4x + 4\]
\[x^{2} - 3x - 40 = 0\]
\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 40\]
\[x_{1} = 8;\ \ \ x_{2} = - 5 < 2.\]
\[Ответ:x = 8.\]