# How do you solve by completing the square: x^2 + 6x + 1 = 0?

x= -3+2$\sqrt{2}$, -3-2$\sqrt{2}$
Rewrite as ${x}^{2}$ +6x+9-9+1=0
${x}^{2}$ +6x+9=8
${\left(x + 3\right)}^{2}$= 8
x+3= 2$\sqrt{2}$, -2$\sqrt{2}$
x= -3 + 2$\sqrt{2}$; -3 -2$\sqrt{2}$