Answer
The solutions are $x=-4\pm\sqrt{15}$
Work Step by Step
$x^{2}+8x+1=0$
Take $1$ to the right side:
$x^{2}+8x=-1$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=8$:
$x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}=-1+\Big(\dfrac{8}{2}\Big)^{2}$
$x^{2}+8x+16=-1+16$
$x^{2}+8x+16=15$
Factor the left side of the equation, which is a perfect square trinomial:
$(x+4)^{2}=15$
Take the square root of both sides:
$\sqrt{(x+4)^{2}}=\pm\sqrt{15}$
$x+4=\pm\sqrt{15}$
Solve for $x$:
$x=-4\pm\sqrt{15}$