x² + y² + 2y - 3 = 0

To solve the equation \( x^2 + y^2 + 2y - 3 = 0 \), we can rearrange and complete the square for \( y \).

First, let's rewrite the equation:

\[ x^2 + y^2 + 2y = 3 \]

Next, we complete the square for the \( y \) term. The expression \( y^2 + 2y \) can be rewritten as follows:

\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting this back into the equation gives us:

\[ x^2 + (y + 1)^2 - 1 = 3 \]

Now, simplify the equation:

\[ x^2 + (y + 1)^2 = 4 \]

This equation describes a circle with center at \( (0, -1) \) and a radius of \( 2 \) (since \( 4 = 2^2 \)).

If you need to find specific points \( (x, y) \) on the circle or anything else related to this equation, just let me know!