18 балов ............

Решите задачу:

$\left \{ {{x+y=-0.2} \atop {x^2+y^2=0.52}} \right. \\ \\ \left \{ {{5x+5y=-1} \atop {25x^2+25y^2=13}} \right. \\ \\ x= \frac{-1-5y}{5} \\ \\ \frac{25(1+5y)^2}{25} +25y^2=13 \\ \\ 1+10y+25y^2+25y^2-13=0 \\ \\ 50y^2+10y-12=0 \\ \\ D=100+4*50*12=2500=50^2 \\ \\ y_{1} =(-10-50)/100=-0.6 \\ x_{1} =0.4 \\ \\ y_{2} =(-10+50)/100=0.4 \\ x_{2} =-0.6 \\ \\OTVET:(0.4;-0.6)(-0.6;0.4) \\ \\$