Answer
$(-1 + \sqrt 3, 1 +\sqrt3)$, $(-1 - \sqrt 3, 1 - \sqrt3)$
Work Step by Step
$x^{2} + y^{2} = 8$
$y = x+2$
Substitute x+2 for y
$x^{2} + (x+2)^{2} = 8$
$2x^{2} + 4x + 4 = 8$
$2x^{2} + 4x = 4 $
$2 (x^{2} + 2x + 1) = 6$
$ (x+1)^{2} = 3$
$(x+1) = +/- \sqrt 3$
Thus $x = -1 +/- \sqrt 3$
So the solutions are $(-1 + \sqrt 3, 1 +\sqrt3)$, $(-1 - \sqrt 3, 1 - \sqrt3)$
