Solve the equation by completing the square. \( 2 x^{2}-x-6=0 \)

To solve \( 2x^2 - x - 6 = 0 \) by completing the square, first divide the entire equation by 2 to make the coefficient of \( x^2 \) equal to 1: \[ x^2 - \frac{1}{2}x - 3 = 0. \] Next, move the constant term to the right: \[ x^2 - \frac{1}{2}x = 3. \] Now, complete the square for the \( x \) terms. Take half of the coefficient of \( x \) (which is \( -\frac{1}{2} \)), square it (giving \( \left(-\frac{1}{4}\right)^2 = \frac{1}{16} \)), and add it to both sides: \[ x^2 - \frac{1}{2}x + \frac{1}{16} = 3 + \frac{1}{16}. \] This simplifies to: \[ \left(x - \frac{1}{4}\right)^2 = \frac{48}{16} + \frac{1}{16} = \frac{49}{16}. \] Now, take the square root of both sides: \[ x - \frac{1}{4} = \pm \frac{7}{4}. \] Finally, solve for \( x \): 1. \( x - \frac{1}{4} = \frac{7}{4} \) leads to \( x = 2 \). 2. \( x - \frac{1}{4} = -\frac{7}{4} \) leads to \( x = -\frac{3}{2} \). Thus, the solutions are \( x = 2 \) and \( x = -\frac{3}{2} \).