Решите систему уравнений методом подстановки {x^2+2y=6,{y=x-1

$\left \{ {{x^2+2y=6} \atop {y=x-1}} \right.\Rightarrow\\\\ x^2+2(x-1)=6\\ x^2+2x-2-6=0\\ x^2+2x-8=0\\ D=4+32=36; \sqrt{D}=6\\\\ x_{1/2}= \frac{-2\pm6}{2}\\\\ x_1=-4; \ x_2=2$

При $x=-4$ :

$(-4)^2+2y=6\\ 2y=6-16\\ y=-10:2\\ y=5\\\\$

При $x=2$

$y=2-1\\ y=1$

Проверка:

$\left \{ {{x^2+2y=6} \atop {y=x-1}} \right.\Rightarrow\left \{ {{(-4)^2+2\cdot(-5)=6} \atop {-5=(-4)-1}} \right.\Rightarrow \left \{ {{16-10=6} \atop {-5=-5}} \right. \left \{ {{6=6} \atop {-5=-5}} \right.\\\\ \left \{ {{x^2+2y=6} \atop {y=x-1}} \right.\Rightarrow\left \{ {{2^2+2\cdot1=6} \atop {1=2-1}} \right.\Rightarrow \left \{ {{6=6} \atop {1=1}} \right.$

Ответ: $x_1=-4, y_1=-5;\ x_2=2, y_2=1$