Solve the inequality \( -2 x^{2}-6 x-4<0 \) The solution has form: \[ \mathrm{A}<\mathrm{x}<\mathrm{B} \] Where \( \mathrm{A}=\square \) or \( \mathrm{x}>\mathrm{B} \) and \( \mathrm{B}=\square \)

To solve the inequality \( -2x^2 - 6x - 4 < 0 \), we first rewrite it in a more standard form by multiplying through by -1 (which reverses the inequality): \[ 2x^2 + 6x + 4 > 0 \] Next, we can simplify this expression by dividing all terms by 2: \[ x^2 + 3x + 2 > 0 \] Now, we will factor the quadratic expression: \[ x^2 + 3x + 2 = (x + 1)(x + 2) \] Thus, we need to solve the inequality: \[ (x + 1)(x + 2) > 0 \] To find the critical points, we set the factors equal to zero: \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] These critical points divide the number line into intervals. We will test the sign of the product \( (x + 1)(x + 2) \) in each of the intervals: 1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \): \[ (-3 + 1)(-3 + 2) = (-2)(-1) = 2 > 0 \] 2. **Interval \( (-2, -1) \)**: Choose \( x = -1.5 \): \[ (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0 \] 3. **Interval \( (-1, \infty) \)**: Choose \( x = 0 \): \[ (0 + 1)(0 + 2) = (1)(2) = 2 > 0 \] Now we summarize the results: - The product \( (x + 1)(x + 2) > 0 \) in the intervals \( (-\infty, -2) \) and \( (-1, \infty) \). - The product \( (x + 1)(x + 2) < 0 \) in the interval \( (-2, -1) \). Thus, the solution to the inequality \( -2x^2 - 6x - 4 < 0 \) is: \[ -2 < x < -1 \] In the required form, we have: \[ \mathrm{A} = -2 \quad \text{and} \quad \mathrm{B} = -1 \] So the final answer is: \[ \mathrm{A} = -2, \quad \mathrm{B} = -1 \]