Решить систему уравнений

2x^2-y^2=32

2x-y=8

Y = 2X - 8

2X^2 - (2X - 8) ^2 = 32

2X^2 - (4X^2 - 32X + 64) = 32

2X^2 - 4X^2 + 32X - 64 - 32 = 0

-2X^2 + 32X - 96 = 0

2 * ( - X^2 + 16X - 48) = 0

D = 256 - 4 * (-1) * (-48) = 256 - 192 = 64; V D = 8

X1 = ( - 16 + 8) : (-2) = + 4

X2 = (24) : ( - 2) = + 12

Y = 2X - 8

Y1 = 2*X1 - 8 = 8 - 8 = 0

Y2 = 2*X2 - 8 = 24 - 8 = 16

Ответ: (4; 0); (12; 16)