1. The problem is to solve the equation $$(x + 4)^2 = -4$$ for all values of $x$. 2. Recall that the square of any real number is always non-negative, so $$(x + 4)^2 \geq 0$$ for all real $x$. 3. Since the right side of the equation is $-4$, which is negative, there are no real solutions to this equation. 4. However, if we consider complex numbers, we can solve it by taking the square root of both sides: $$ (x + 4)^2 = -4 $$ $$ x + 4 = \pm \sqrt{-4} $$ 5. Using the imaginary unit $i$ where $i^2 = -1$, we have: $$ \sqrt{-4} = \sqrt{4 \times (-1)} = \sqrt{4} \times \sqrt{-1} = 2i $$ 6. So, $$ x + 4 = \pm 2i $$ 7. Finally, solve for $x$ by subtracting 4 from both sides: $$ x = -4 \pm 2i $$ This gives two complex solutions: $$ x = -4 + 2i \quad \text{and} \quad x = -4 - 2i $$