Решите уравнение (х+2)*√(23-4х)-3х^2=0

[latex](x+2)* \sqrt{(23-4x)-3x^2} =0[/latex] ОДЗ: [latex]23-4x-3x^2 \geq 0[/latex] [latex]3 x^{2} +4x-23 \leq 0[/latex] [latex]D=16+276=292[/latex] [latex]x_1= \frac{-2+ \sqrt{73} }{3} [/latex] [latex]x_2= \frac{-2- \sqrt{73} }{3} [/latex] [latex]x[/latex]∈[latex][ \frac{-2- \sqrt{73} }{3} ; \frac{-2+ \sqrt{73} }{3} ][/latex] [latex]x+2=0[/latex]   или [latex] \sqrt{23-4x-3x^2} =0[/latex] [latex]x=-2[/latex]  или [latex]x_1= \frac{-2+ \sqrt{73} }{3} [/latex]  [latex]x_2= \frac{-2- \sqrt{73}}{3} [/latex] Ответ: [latex]-2; \frac{-2+ \sqrt{73} }{3} ; \frac{-2- \sqrt{73} }{3} [/latex]

[latex]23 -4x -3x^2 \geq 0; \ \ \ 3x^2 +4x-23 \leq 0 \\ \\ 3x^2+4x-23=0; \ \ x_{1,2}=\frac{b^2 \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a}=\frac{-4 \pm \sqrt{16 +276}}{6}=\frac{-4 \pm 2 \sqrt{73}}{6} \\ \\ x_1= \frac{\sqrt{73}-2}{3}; \ \ x_2=\frac{-2 - \sqrt{73}}{3} \\ \\ x \in [\frac{-2 - \sqrt{73}}{3}}; \frac{\sqrt{73}-2}{3}][/latex] [latex]x+2 =0; \ \ x=-2 \in [\frac{-2 - \sqrt{73}}{3}}; \frac{\sqrt{73}-2}{3}] \\ \\ OTBET: \ -2; \ \frac{-2 - \sqrt{73}}{3}}, \ \frac{\sqrt{73}-2}{3}[/latex]