The problem requires solving the quadratic equation $x^2 - 3x - 5 = 0$ using the completing the square method. Let’s solve it step by step:

Step 1: Rearrange the equation

Bring the constant term ($-5$) to the other side: $x^2 - 3x = 5$

Step 2: Complete the square

Take half the coefficient of $x$ (which is $-3$), square it, and add it to both sides: $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$ Add $\frac{9}{4}$ to both sides: $x^2 - 3x + \frac{9}{4} = 5 + \frac{9}{4}$

Step 3: Simplify both sides

The left-hand side becomes a perfect square trinomial: $\left(x - \frac{3}{2}\right)^2 = \frac{20}{4} + \frac{9}{4} = \frac{29}{4}$

Step 4: Solve for $x$

Take the square root of both sides: $x - \frac{3}{2} = \pm \sqrt{\frac{29}{4}}$ Simplify the square root: $x - \frac{3}{2} = \pm \frac{\sqrt{29}}{2}$

Step 5: Isolate $x$

Add $\frac{3}{2}$ to both sides: $x = \frac{3}{2} \pm \frac{\sqrt{29}}{2}$

Combine the terms into a single fraction: $x = \frac{3 \pm \sqrt{29}}{2}$

Final Answer:

The solutions are: $x = \frac{3 + \sqrt{29}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{29}}{2}.$

Would you like further clarification, or do you have additional questions?

Here are some related questions for practice:

1. How do you apply completing the square when the coefficient of $x^2$ is not 1?
2. How does completing the square relate to the quadratic formula?
3. What is the geometric meaning of completing the square?
4. Can you derive the vertex form of a quadratic equation using this method?
5. How does adding and subtracting terms affect equality in an equation?

Tip: Always ensure the coefficient of $x^2$ is 1 before starting the completing-the-square process.