## Algebra and Trigonometry 10th Edition

{$-4-\sqrt 2, -4+\sqrt 2$}
$x^2+8x+14=0$ $x^2+8x+16=2$ $(x+4)^2=2$ $x+4=+\sqrt 2 or -\sqrt 2$ x = {$-4-\sqrt 2, -4+\sqrt 2$}