Solve by completing the square:
[tex]\begin{gathered} x^2+6x+8=0 \\ x^2+6x=-8 \end{gathered}[/tex]We have to find the third term for x^2+6x in order to complete a binomial (x+3)^2:
[tex](x+3)^2=x^2+2\cdot3x+9=x^2+6x+9[/tex]So we have to add 9:
[tex]\begin{gathered} x^2+6x+9=-8+9 \\ (x+3)^2=1 \end{gathered}[/tex]Then, we apply the square root:
[tex]\begin{gathered} \sqrt[]{(x+3)^2}=\sqrt[]{1} \\ x+3=\pm1 \end{gathered}[/tex][tex]\begin{gathered} x_1+3=-1 \\ x_1=-1-3=-4 \end{gathered}[/tex][tex]\begin{gathered} x_2+3=+1 \\ x_2=1-3=-2 \end{gathered}[/tex]Answer: The solutions are x=-4 and x=-2
