- 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:

\[ x \in (-\infty, 0] \cup \left[\frac{3}{2}, \infty\right) \]