Solve the following equation for $y$ and sketch a grapl Round to two decimal places where necessary. \[ x=y^{2}+2 y-3 \] \[ y=\square \pm \sqrt{x-} \]

Step 1 ：The given equation is a quadratic equation in terms of y. We can solve for y using the quadratic formula. The quadratic formula is given by: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where a, b, and c are the coefficients of the quadratic equation. In this case, a = 1, b = 2, and c = -x-3. We can substitute these values into the quadratic formula to solve for y.

Step 2 ：The solutions to the equation are \[y = -\sqrt{x + 4} - 1\] and \[y = \sqrt{x + 4} - 1\]. These are the two branches of the parabola defined by the original equation.

Step 3 ：The graph of these solutions is a parabola that opens to the right if y is considered as the dependent variable and x as the independent variable. The vertex of the parabola is at (-4,-1) and the parabola is symmetric about the line y=-1.

Step 4 ：\(\boxed{\text{Final Answer: The solutions to the equation are } y = -\sqrt{x + 4} - 1 \text{ and } y = \sqrt{x + 4} - 1. \text{ The graph of these solutions is a parabola that opens to the right if y is considered as the dependent variable and x as the independent variable. The vertex of the parabola is at } (-4,-1) \text{ and the parabola is symmetric about the line } y=-1.}\)