Solve the Quadratic Equation with Negative Coefficient: -(x+3)²=4x

To solve $-(x+3)^2 = 4x$, follow these steps:

• Step 1: Expand the left side: $-(x+3)^2 = -(x^2 + 6x + 9)$.
• Step 2: Distribute the negative sign: $-x^2 - 6x - 9$.
• Step 3: Set the equation by moving terms to the right: $-x^2 - 6x - 9 = 4x$ becomes $-x^2 - 6x - 9 - 4x = 0$.
• Step 4: Simplify to standard quadratic form: $-x^2 - 10x - 9 = 0$.
• Step 5: Applying the quadratic formula where $a = -1$, $b = -10$, $c = -9$:
• $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot (-1) \cdot (-9)}}{2 \cdot (-1)}$.
• $x = \frac{10 \pm \sqrt{100 - 36}}{-2}$.
• $x = \frac{10 \pm \sqrt{64}}{-2}$.
• $x = \frac{10 \pm 8}{-2}$.
• Two solutions arise: $x = \frac{10 + 8}{-2} = -9$ and $x = \frac{10 - 8}{-2} = -1$.

Therefore, the solutions are $x_1 = -1$ and $x_2 = -9$.

Thus, the correct answer is $\mathbf{x_1=-1,x_2=-9}$.