Решить уравнение √(4-x)-√(5+x)=3

[latex]\sqrt{4-x}-\sqrt{5+x}=3\\O.D.3.:\\\begin{cases}4-x\geq0\\5+x\geq0\end{cases}\Rightarrow\begin{cases}x\leq4\\x\geq-5\end{cases}\Rightarrow x\in[-5;\;4]\\\sqrt{4-x}=\sqrt{5+x}+3\\4-x=5+x+6\sqrt{5+x}+9\\6\sqrt{5+x}=4-x-5-x-9\\6\sqrt{5+x}=-10-2x\\3\sqrt{5+x}=-5-x\\9(5+x)=25+10x+x^2\\x^2+10x+25-45-9x=0\\x^2+x-20=0\\D=1+4\cdot20=81\\x_{1,2}=\frac{-1\pm9}2\\x_1=-5,\;x_2=4[/latex]