## College Algebra (10th Edition)

The solution set is {$-2-\sqrt 2, -2+\sqrt 2$}.
$x^2+4x+2=0$ $a=1, b=4, c=2$ $D=b^2-4ac=(4)^2-4\cdot1\cdot2=16-8=8>0$ The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-4\pm\sqrt 8}{2}$ Use quadratic formula. $x=\frac{-4\pm2\sqrt 2}{2}$ Simplify $x=-2\pm\sqrt 2$ $x=-2+\sqrt 2$, $x=-2-\sqrt 2$ The solution set is {$-2-\sqrt 2, -2+\sqrt 2$}