Verified answer

[tex]\displaystyle\bf\\\left \{ {{x^{2} +y^{2}=9 } \atop {y=x^{2} -3}} \right. \\\\\\\left \{ {{x^{2} =y+3} \atop {y+3+y^{2} =9}} \right.\\\\\\\left \{ {{x^{2} =y+3} \atop {y^{2} +y-6=0}} \right. \\\\\\y^{2} +y-6=0\\\\D=1^{2} -4\cdot(-6)=1+24=25=5^{2} \\\\\\y_{1} =\frac{-1+5}{2} =\frac{4}{2} =2\\\\\\y_{2} =\frac{-1-5}{2} =\frac{-6}{2}=-3 \\\\\\1)\\\\y_{1} =2\\\\x^{2} =y+3=2+3=5\\\\x_{1} =-\sqrt{5} \ \ ; \ \ x_{2} =\sqrt{5} \\\\2)\\\\y_{2} =-3\\\\x^{2} =y+3=-3+3=0\\\\x=0[/tex]

Ответ : три решения   [tex]\displaystyle\bf\\\Big(-\sqrt{5} \ ; \ 2\Big) \ , \ \Big(\sqrt{5} \ , \ 2\Big) \ , \ \Big(0 \ ; \ -3\Big)[/tex]