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.

Learn how to solve the quadratic equation y^2 + 1 = 3 with a detailed step-by-step solution. This problem covers key algebraic concepts, including solving quadratic equations using the square root property. Suitable for students in Grades 7-9.

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