Решите систему урaвнений -x+y=-6-4y-x^2-3

$\left \{ {{-x+y=-6} \atop {-4y- x^{2} =3}} \right.$
$\left \{ {{y=-6+x} \atop {-4(-6=x)- x^{2} =3}} \right.$
$-4(-6+x)- x^{2} =3$
$24-4x- x^{2} -3=0$
$x^{2} +4x-21=0$
$D= 4^{2} -4*(-21)=16+84=100$
$x1= \frac{-4+ \sqrt{100} }{2} = \frac{-4+10}{2} =3$
$x2= \frac{-4-10}{2} = \frac{-14}{2} =-7$

$\left \{ {{x=3} \atop {y=-3}} \right.$
$\left \{ {{x=-7} \atop {y=-13}} \right.$
Ответ:(3;-3) и (-7;-13)