Х-у=6 ху=8 в фигурной скобке

Решите задачу:

$\displaystyle \left \{ {{x-y=6} \atop {xy=8}} \right. \\ \\ \left \{ {{x=y+6} \atop {(y+6)y=8}} \right. \\ \\ y^{2}+6y-8=0 \\ D=b^{2}-4ac=36+32=68=( 2\sqrt{17})^{2} \\ \\ y_{1}= \frac{-b+ \sqrt{D} }{2a}= \frac{-6+2 \sqrt{17} }{2}=-3+\sqrt{17} \\ \\ y_{2}= \frac{-b- \sqrt{D} }{2a}= \frac{-6-2 \sqrt{17} }{2}=-3- \sqrt{17} \\ \\ x_{1}=6-3+ \sqrt{17}=3+ \sqrt{17} \\ x_{2} =6-3- \sqrt{17} =3- \sqrt{17}$