Решить систему уравнений :2х2-у2=32,2х-у=8

$\left \{ {{2 x^{2} - y^{2} =32} \atop {2x-y=8}} \right. \\ \\ \left \{ {{y=2x-8} \atop {(2 x^{2} -(2x-8)^{2}=32}} \right. \\ \\ \left \{ {{y=2x-8} \atop {2 x^{2} -4 x^{2} +32x-64=32}} \right. \\ \\ \left \{ {{y=2x-8} \atop {-2 x^{2} +32x-96=0}} \right. \\ \\ 2 x^{2} -32x+96=0 \\ x^{2} -16x+48=0 \\ \\ \left \{ {{x_{1}+x_{2}=16} \atop {x_{1}x_{2}=48}} \right.$
$x_{1}=12 \\ x_{2}=4$

Ответ: $12; 4$