Jak rozwiązać taki układ równań? x+y =6 xy =4

[latex]egin{cases}x+y =6 \ xy =4 end{cases}\ \ egin{cases}x =-y+6 \ y (-y+6 )=4 end{cases}\ \ egin{cases}x =-y+6 \ -y^2+6y-4 =0end{cases}\ \Delta = b^{2}-4ac = 6^{2}-4*(-1)* (-4)= 36-16=20[/latex] [latex]sqrt{Delta }=sqrt{20}=sqrt{4*5}=2sqrt{5}\ \y_{1}=frac{-b-sqrt{Delta }}{2a} =frac{-6-2sqrt{5}}{-2}=frac{-2( 3+sqrt{5})}{-2}=3+sqrt{5}\ \y_{2}=frac{-b+sqrt{Delta }}{2a} =frac{-6+2sqrt{5}}{-2}=frac{-2( 3-sqrt{5})}{-2}=3-sqrt{5}[/latex] [latex]egin{cases}x =-(3+sqrt{5})+6 \ y=3+sqrt{5}end{cases}\ \egin{cases}x =-3-sqrt{5} +6 \ y=3+sqrt{5}end{cases}\ \egin{cases}x = 3-sqrt{5} \ y=3+sqrt{5}end{cases}\ \ lub \ \[/latex] [latex]egin{cases}x =-(3-sqrt{5})+6 \ y=3-sqrt{5}end{cases}\ \egin{cases}x =-3+sqrt{5} +6 \ y=3-sqrt{5}end{cases}\ \egin{cases}x = 3+sqrt{5} \ y=3-sqrt{5}end{cases}[/latex]