Решите систему уравнений {-x-y=3 {x^2+y^2=5

$\left \{ {{-x-y=3} \atop { x^{2} + y^{2}=5 }} \right.$ ⇒ $\left \{ {{x=-3-y} \atop { (-3-y)^{2} + y^{2} =5 }} \right.$ ⇒ $\left \{ {{x=-3-y} \atop {9+6y+2 y^{2}=5 }} \right.$ ⇒ $\left \{ {{x=-3-y} \atop { y^{2} +3y+2=0 }} \right.$
D=9-8=1
y(1)=(-3+1)/2=-1
y(2)= -2
x(1)=-3-y(1)=-3+1=-2
x(2)=-3-y(2)=-3+2=-1
ответ; (-2;-1), (-1;-2)