Решите систему уравнений заранее спасибо

Решите задачу:

$\left \{ {{ x^{2} -3 y^{2} =24} \atop {x+y=8}} \right. \left \{ {{(8-y)- 3y^{2} =24} \atop {x=8-y}} \right. \\ (8-y)-3 y^{2} -24=0 \\ 64-16y+ y^{2} -3 y^{2} -24=0 \\ 40-16y-2 y^{2} =0 \\ 20-8y- y^{2} =0 \\ D=8 ^{2} -4*20*(-1)=144= x=12^{2} \\ y_{1} =0.5 \\ y_{2} =-0.1 \\ x_{1} -8-0.5=7.5 \\ x_{2} =8-(-0.1)=8.1$