Solve 6x² + 3x − 1 = 0 using completing the square.

$$\frac{6x^2}{6} + \frac{3x}{6} - \frac{1}{6} = \frac{0}{6}$$ $$x^2 + \frac{1}{2}x - \frac{1}{6} = 0$$

Take half of the coefficient of $x$ ($\frac{1}{2}$), square it ($\left(\frac{1}{4}\right)^2 = \frac{1}{16}$), and add it to both sides of the equation.

$$x^2 + \frac{1}{2}x + \frac{1}{16} = \frac{1}{6} + \frac{1}{16}$$

$$\left(x + \frac{1}{4}\right)^2 = \frac{8}{48} + \frac{3}{48}$$ $$\left(x + \frac{1}{4}\right)^2 = \frac{11}{48}$$

$$x + \frac{1}{4} = \pm\sqrt{\frac{11}{48}}$$ $$x + \frac{1}{4} = \pm\frac{\sqrt{11}}{\sqrt{48}}$$

Simplify $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$.

$$x + \frac{1}{4} = \pm\frac{\sqrt{11}}{4\sqrt{3}}$$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$$x + \frac{1}{4} = \pm\frac{\sqrt{11}\sqrt{3}}{4\sqrt{3}\sqrt{3}}$$ $$x + \frac{1}{4} = \pm\frac{\sqrt{33}}{12}$$

$$x = -\frac{1}{4} \pm \frac{\sqrt{33}}{12}$$

Combine the terms with a common denominator of 12:

$$x = -\frac{3}{12} \pm \frac{\sqrt{33}}{12}$$ $$x = \frac{-3 \pm \sqrt{33}}{12}$$