Solve $x^{2}+6 x+7=0$
\[
x=-1 \text { and } x=-5
\]
$3 \pm \sqrt{2}$
$-3 \pm \sqrt{2}$
$\frac{-3 \pm \sqrt{2}}{2}$

Answer

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Answer

So, the solutions to the equation $x^{2}+6 x+7=0$ are $\boxed{x = -3 + \sqrt{2}}$ and $\boxed{x = -3 - \sqrt{2}}$

Steps

Step 1 ：$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Step 2 ：Here, $a=1$, $b=6$, and $c=7$

Step 3 ：Substituting these values into the quadratic formula, we get:

Step 4 ：$x = \frac{-6 \pm \sqrt{6^{2}-4*1*7}}{2*1}$

Step 5 ：$x = \frac{-6 \pm \sqrt{36-28}}{2}$

Step 6 ：$x = \frac{-6 \pm \sqrt{8}}{2}$

Step 7 ：$x = \frac{-6 \pm 2\sqrt{2}}{2}$

Step 8 ：$x = -3 \pm \sqrt{2}$

Step 9 ：So, the solutions to the equation $x^{2}+6 x+7=0$ are $\boxed{x = -3 + \sqrt{2}}$ and $\boxed{x = -3 - \sqrt{2}}$