Помогите: X^2+y=12 Знак системы X+y^2=12

Решите задачу:

$x^2+y=12\\ x+y^2=12\\\\ x^2+y=x+y^2\\ x^2-y^2=x-y\\ (x+y)(x-y)=x-y\\ (x-y)(x+y-1)=0\\ x=y\\ x^2+x-12=0\\ D=1+4*1*12=7^2\\ x=\frac{-1+7}{2}=3\\ x=\frac{-1-7}{2}=-4\\ y=3\\ y=-4\\ \\ x=1-y\\ (1-y)^2+y=12\\ 1-2y+2y^2=12\\ 2y^2-2y-11=0\\ D=4+4*2*11=\sqrt{92}\\ y=\frac{2+\sqrt{92}}{4}\\ y=\frac{2-\sqrt{92}}{4}\\ x=\frac{2-\sqrt{92}}{4}\\ x=\frac{2+\sqrt{92}}{4}$