Your equation is already a perfect square, so we don't have to complete anything:
[tex]x^2+8x+16 = (x+4)^2[/tex]
So, we can write the equation as
[tex](x+4)^2=2[/tex]
Which is true if and only if
[tex]x+4 = \sqrt{2} \lor x+4=-\sqrt{2} \iff x = \sqrt{2}-4 \lor x = -2-\sqrt{4}[/tex]
