Divide the equation by the coefficient of x², which is 2.

$$2x^2 + 4x + 1 = 0$$ $$\frac{2x^2}{2} + \frac{4x}{2} + \frac{1}{2} = \frac{0}{2}$$ $$x^2 + 2x + \frac{1}{2} = 0$$

To do this, take half of the coefficient of $x$ (which is 2), square it, and add it to both sides of the equation.
Half of the coefficient of $x$ is $\frac{2}{2} = 1$.
Squaring it gives $1^2 = 1$.

$$x^2 + 2x + 1 = -\frac{1}{2} + 1$$

$$\sqrt{(x+1)^2} = \pm\sqrt{\frac{1}{2}}$$ $$x+1 = \pm\frac{1}{\sqrt{2}}$$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$$x+1 = \pm\frac{\sqrt{2}}{2}$$

$$x = -1 \pm \frac{\sqrt{2}}{2}$$

The two solutions are:

$$x_1 = -1 + \frac{\sqrt{2}}{2}$$ $$x_2 = -1 - \frac{\sqrt{2}}{2}$$

The solutions can also be written as:

$$x = \frac{-2 \pm \sqrt{2}}{2}$$

The final answers are $x = -1 + \frac{\sqrt{2}}{2}, x = -1 - \frac{\sqrt{2}}{2}$.