1. The problem is to analyze the equation $y = (x + y + 1)^2$ and understand its behavior. 2. Start by expanding the right side: $$y = (x + y + 1)^2 = (x + y + 1)(x + y + 1) = x^2 + 2xy + y^2 + 2x + 2y + 1$$ 3. Rearrange the equation to isolate terms: $$y = x^2 + 2xy + y^2 + 2x + 2y + 1$$ 4. Move all terms to one side to set the equation to zero: $$0 = x^2 + 2xy + y^2 + 2x + 2y + 1 - y$$ 5. Simplify the $y$ terms: $$0 = x^2 + 2xy + y^2 + 2x + y + 1$$ 6. This is a quadratic equation in terms of $y$: $$y^2 + (2x + 1)y + (x^2 + 2x + 1) = 0$$ 7. To solve for $y$, use the quadratic formula: $$y = \frac{-(2x + 1) \pm \sqrt{(2x + 1)^2 - 4(x^2 + 2x + 1)}}{2}$$ 8. Calculate the discriminant: $$(2x + 1)^2 - 4(x^2 + 2x + 1) = 4x^2 + 4x + 1 - 4x^2 - 8x - 4 = -4x - 3$$ 9. The discriminant is $-4x - 3$, which determines the nature of the roots. 10. The solutions for $y$ are: $$y = \frac{-(2x + 1) \pm \sqrt{-4x - 3}}{2}$$ Final answer: The equation defines $y$ implicitly with solutions depending on $x$ and the discriminant $-4x - 3$. Real solutions exist only when $-4x - 3 \geq 0$, i.e., $x \leq -\frac{3}{4}$.