- 4x ^ 2 + 6x <= 0

To solve the inequality \( -4x^2 + 6x \leq 0 \), we can first rewrite it in a standard form: \[ 4x^2 - 6x \geq 0 \] Next, we can factor the left side: \[ 2x(2x - 3) \geq 0 \] Now we will determine the values of \( x \) for which this inequality holds. We first find the critical points by setting the equation to zero: \[ 2x(2x - 3) = 0 \] This gives us: 1. \( 2x = 0 \) → \( x = 0 \) 2. \( 2x - 3 = 0 \) → \( x = \frac{3}{2} \) The critical points that divide the number line into intervals are \( x = 0 \) and \( x = \frac{3}{2} \). We will test values in the intervals created by these points: \( (-\infty, 0) \), \( (0, \frac{3}{2}) \), and \( (\frac{3}{2}, \infty) \). 1. **Interval \( (-\infty, 0) \)**: Pick \( x = -1 \) \[ 2(-1)(2(-1) - 3) = 2(-1)(-2 - 3) = 2(-1)(-5) = 10 \geq 0 \quad \text{(True)} \] 2. **Interval \( (0, \frac{3}{2}) \)**: Pick \( x = 1 \) \[ 2(1)(2(1) - 3) = 2(1)(2 - 3) = 2(1)(-1) = -2 \geq 0 \quad \text{(False)} \] 3. **Interval \( (\frac{3}{2}, \infty) \)**: Pick \( x = 2 \) \[ 2(2)(2(2) - 3) = 2(2)(4 - 3) = 2(2)(1) = 4 \geq 0 \quad \text{(True)} \] Now we compile our findings: - The inequality is satisfied in the intervals \( (-\infty, 0] \) and \( [\frac{3}{2}, \infty) \). Thus, the solution to the inequality \( -4x^2 + 6x \leq 0 \) is: \[ x \in (-\infty, 0] \cup \left[\frac{3}{2}, \infty\right) \]