Answer
See below
Work Step by Step
Given: $x^2-2y=6\\x^2-y^2=-27$
Multiply the second equation by $-1$ and add it to the second equation:
$$y^2-2y=33\\y^2-2y-33=0$$
Solve for this: $y=\frac{-b \pm \sqrt b^2-4ac}{2a}\\=\frac{-(-2)\pm\sqrt (-2)^2-4(1)(-33)}{2(1)}\\=\frac{2\pm \sqrt 136}{2}\\=1\pm \sqrt 34$
Substitute back to $y$: $x^2-2(1+\sqrt 34)=6\\x^2=6+2(1+\sqrt 34)\\x^2=8+2\sqrt 34\\\rightarrow x=\pm \sqrt 8+2\sqrt 34$
$x^2-2(1-\sqrt 34)=6\\x^2=6+2(1-\sqrt 34)\\x^2=8-2\sqrt 34\\\rightarrow x=\pm \sqrt 8-2\sqrt 34$
