X^2+3xy+9y^2=12 и x^2+3xy+2y^2=0 система пожалуйста дам много баллов!!!!!!!!!!!!!!!

$\left \{ {{x^2+3xy+9y^2=12} \atop {x^2+3xy+2y^2=0}} \right. \; ,\; \left \{ {{(x^2+3xy+2y^2)+7y^2=12} \atop {x^2+3xy+2y^2=0}} \right. \; ,\; \left \{ {{7y^2=12} \atop {x^2+3xy+2y^2=0}} \right. \\\\ \left \{ {{y=\pm \sqrt{\frac{12}{7}}} \atop {x^2+3xy+2y^2=0}} \right. \; \Rightarrow \\\\1)\; \; \left \{ {{y=+\sqrt{\frac{12}{7}}} \atop {x^2+3\cdot \sqrt{\frac{12}{7}}\cdot x+2\cdot \frac{12}{7}=0}} \right. \\\\D=9\cdot \frac{12}{7}-4\cdot \frac{2\cdot 12}{7}=\frac{12}{7}$

$x_{1,2}=\frac{-3\sqrt{\frac{12}{7}}\pm \sqrt{\frac{12}{7}}}{2}\; ;\; \; x_1=-2\sqrt{\frac{12}{7}}\; ,\; \; x_2=-\sqrt{\frac{12}{7}}\\\\2) \left \{ {{y=-\sqrt{\frac{12}{7}}} \atop {x^2-3\cdot \sqrt{\frac{12}{7}}+2\cdot \frac{12}{7}=0}} \right. \; ,\\\\D=9\cdot \frac{12}{7}-8\cdot \frac{12}{7}= \frac{12}{7}\\\\x_1=\frac{3\sqrt{\frac{12}{7}}-\sqrt{\frac{12}{7}}}{2} =\sqrt{\frac{12}{7}}\; ,\\\\ x_2= 2\sqrt{\frac{12}{7}}\; .$

Ответ: $(-\sqrt{\frac{12}{7}},\sqrt{\frac{12}{7}});(-2\sqrt{\frac{12}{7}},\sqrt{\frac{12}{7}});(\sqrt{\frac{12}{7}},-\sqrt{\frac{12}{7}}),(2\sqrt{\frac{12}{7}},-\sqrt{\frac{12}{7}}).$