Answer
$x=-2\pm\sqrt{\dfrac{7}{2}}$
Work Step by Step
$2x^{2}+8x+1=0$
Let's take the independent term to the right side of the equation:
$2x^{2}+8x=-1$
Take out common factor $2$ on the left side of the equation:
$2(x^{2}+4x)=-1$
Let's complete the square. Remember that, in order to do that, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation $b=4$
Please note that, since the expression whose square needs to be completed is multiplied by $2$, we will add $2(\dfrac{b}{2})^{2}$ to the right side of the equation.
$2[x^{2}+4x+(\dfrac{4}{2})^{2}]=-1+2(\dfrac{4}{2})^{2}$
$2(x^{2}+4x+4)=-1+8$
$2(x^{2}+4x+4)=7$
$x^{2}+4x+4=\dfrac{7}{2}$
We have a perfect square trinomial on the left side of the expression. We factor it and the equation becomes:
$(x+2)^{2}=\dfrac{7}{2}$
Take the square root of both sides of the equation:
$\sqrt{(x+2)^{2}}=\sqrt{\dfrac{7}{2}}$
$x+2=\pm\sqrt{\dfrac{7}{2}}$
Solve for $x$:
$x=-2\pm\sqrt{\dfrac{7}{2}}$
