Решите систему уравнений x^2-17y=-138, х+y=4.

$\left \{ {{ x^{2} -17y=-138} \atop {y=4-x}} \right. \left \{ {{x^{2} -17(4-x)=-138} \atop {y=4-x}} \right. \left \{ {x^{2} -68+17x=-138} \atop {y=4-x}} \right. \left \{ {{x^{2} +17x+70=0} \atop {y=4-x}} \right. \\ \\ x^{2} +17x+70=0 \\ D= 17^{2} -4*1*70=289-280=9\ \textgreater \ 0; \sqrt{D} =3 \\ x1= \frac{-17-3}{2} =-10 \\ x2=\frac{-17+3}{2}=-7 \\$
Если x1=-10 и x2=-7, то y1=4+10=14 и y2=4+7=11