To solve the equation $y^2 + 1 = 3$, follow these steps:

1. Subtract 1 from both sides of the equation to isolate the term with $y^2$: $y^2 = 3 - 1$ $y^2 = 2$

2. Take the square root of both sides to solve for $y$: $y = \pm \sqrt{2}$

So, the solutions are $y = \sqrt{2}$ and $y = -\sqrt{2}$.

Would you like more details on the solution or have any other questions?

Here are 5 related questions you might find interesting:

1. How would you solve $y^2 - 4 = 0$?
2. What is the solution to the equation $y^2 + 9 = 25$?
3. Can you solve the equation $y^2 + 1 = 0$ for real numbers?
4. What happens when you solve $y^2 = 4$ by factoring?
5. How do you solve $y^3 + y - 2 = 0$ for $y$?

Tip: When solving quadratic equations, always check if the solution could involve taking square roots of negative numbers, which leads to complex solutions.