To solve the quadratic equation given by
\[ -9 + \frac{(x+3)^2}{-4} = -4, \]
we can start by isolating the term involving \( (x + 3)^2 \).
First, add 9 to both sides of the equation:
\[ \frac{(x + 3)^2}{-4} = -4 + 9 \]
This simplifies to:
\[ \frac{(x + 3)^2}{-4} = 5. \]
Next, multiply both sides of the equation by -4 to eliminate the fraction:
\[ (x + 3)^2 = -20. \]
Since we have a square equal to a negative number, this means that there are no real solutions, because the square of a real number cannot be negative.
However, we can check if the original equation leads to any complex solutions by continuing:
Taking the square root of both sides gives:
\[ x + 3 = \pm \sqrt{-20}. \]
Since \(\sqrt{-20} = \sqrt{20}i = 2\sqrt{5}i\), we can rewrite the equation:
\[ x + 3 = \pm 2\sqrt{5}i. \]
Subtracting 3 from both sides, we obtain:
\[ x = -3 \pm 2\sqrt{5}i. \]
Thus we determine there are complex solutions, but no real solutions exist within the given options. None of the proposed sets of numbers represent the correct solutions to the equation as they are all real. Therefore, based on the options provided, there must be a misunderstanding in the problem statement since the solutions involve complex numbers, not real numbers.
In conclusion, the equation has no real solutions.