Answer
{$-2-\sqrt 2,-2+\sqrt 2$}
Work Step by Step
Step 1: Comparing $y^{2}+4y+2=0$ to the standard form of a quadratic equation $ay^{2}+by+c=0$, we obtain:
$a=1$, $b=4$ and $c=2$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(4) \pm \sqrt {(4)^{2}-4(1)(2)}}{2(1)}$
Step 4: $x=\frac{-4\pm \sqrt {16-8}}{2}$
Step 5: $x=\frac{-4 \pm \sqrt {8}}{2}$
Step 6: $x=\frac{-4 \pm \sqrt {4\times2}}{2}$
Step 7: $x=\frac{-4 \pm \sqrt {2^{2}\times2}}{2}$
Step 8: $x=\frac{-4 \pm 2\sqrt {2}}{2}$
Step 9: $x=-2\pm\sqrt 2$
Step 10: $x=-2+\sqrt 2$ or $x=-2-\sqrt 2$
Step 11: Therefore, the solution set is {$-2-\sqrt 2,-2+\sqrt 2$}.
