To solve the quadratic equation $2x^2 + x - 6 = 0$ by completing the square, follow these steps:

Step 1: Divide the entire equation by the coefficient of $x^2$

The coefficient of $x^2$ is 2. Divide the entire equation by 2:

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

Step 2: Move the constant term to the right side of the equation

Add 3 to both sides of the equation:

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

Step 3: Complete the square

To complete the square, take half of the coefficient of $x$ (which is $\frac{1}{2}$), square it, and add it to both sides of the equation.

Half of $\frac{1}{2}$ is $\frac{1}{4}$, and squaring it gives $\left(\frac{1}{4}\right)^2 = \frac{1}{16}$.

Add $\frac{1}{16}$ to both sides:

$x^2 + \frac{1}{2}x + \frac{1}{16} = 3 + \frac{1}{16}$

Step 4: Simplify the equation

The left side of the equation is now a perfect square trinomial:

$\left(x + \frac{1}{4}\right)^2 = \frac{48}{16} + \frac{1}{16} = \frac{49}{16}$

Step 5: Solve for $x$

Take the square root of both sides:

$x + \frac{1}{4} = \pm \sqrt{\frac{49}{16}}$

Simplify the square root:

$x + \frac{1}{4} = \pm \frac{7}{4}$

Step 6: Isolate $x$

Finally, solve for $x$ by subtracting $\frac{1}{4}$ from both sides:

$x = -\frac{1}{4} \pm \frac{7}{4}$

This gives two solutions:

$x = \frac{6}{4} = \frac{3}{2} \quad \text{or} \quad x = -\frac{8}{4} = -2$

So the solutions are $x = \frac{3}{2}$ and $x = -2$.

Would you like more details or have any questions? Here are five related questions to explore:

1. How do you solve a quadratic equation by factoring?
2. What is the quadratic formula and how is it derived?
3. Can every quadratic equation be solved by completing the square?
4. What is the discriminant, and how does it relate to the number of real solutions?
5. How can you apply the method of completing the square to solve a quadratic inequality?

Tip: Always check your solutions by plugging them back into the original equation to ensure they satisfy it.