Решите неравенство, пожалуйста [latex] \sqrt{x-3} \ \textless \ 5-x[/latex]

[latex] \sqrt{x-3}\ \textless \ 5-x\\ [/latex] ОДЗ: [latex]x-3 \geq 0\\ x \geq 3[/latex] [latex]x\in [3; +\infty)[/latex] [latex] \sqrt{x-3}\ \textless \ 5-x\\ x-3\ \textless \ (5-x)^2 \\ x-3\ \textless \ 25-10x+x^2\\ x^2-10x-x+25+3\ \textless \ 0\\ x^2-11x+28\ \textless \ 0\\ D=121-112=9; \ \sqrt{D}=3\\\\ x_{1/2}= \frac{11\pm3}{2}\\\\ x_1=7\\\\ x_2=4 [/latex] Ответ: [latex]x\in [3; 4) [/latex]