The equation you provided is:

$(x - 1)^2 = 3$

To solve for $x$, follow these steps:

Step 1: Take the square root of both sides

$\sqrt{(x - 1)^2} = \pm \sqrt{3}$

This simplifies to: $x - 1 = \pm \sqrt{3}$

Step 2: Solve for $x$

Now, isolate $x$ by adding 1 to both sides: $x = 1 \pm \sqrt{3}$

So, the solutions are: $x = 1 + \sqrt{3} \quad \text{or} \quad x = 1 - \sqrt{3}$

These are the two possible values of $x$.

Would you like a deeper explanation or have any further questions? Here are some related questions for further practice:

1. What are the approximate decimal values of $1 + \sqrt{3}$ and $1 - \sqrt{3}$?
2. How do you solve equations involving other roots like cube roots?
3. What happens if the equation is $(x - 1)^2 = -3$?
4. How can you factor quadratic expressions like $(x - 1)^2$ when expanded?
5. What are some applications of quadratic equations in physics or engineering?

Tip: Always remember that taking the square root introduces both the positive and negative roots, which is why you have the $\pm$ symbol in the solution.