x

To determine which of the given values is a solution of the quadratic equation \(-2x^2 - 3x + 2 = 0\), we need to substitute each value of \(x\) into the equation and check if it results in zero.

The values to test are:
1. \(x = -\frac{3}{4}\)
2. \(x = -\frac{1}{2}\)
3. \(x = \frac{1}{2}\)

Let's evaluate the equation for each value.

### Testing \(x = -\frac{3}{4}\):
\[
-2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2
\]
\[
= -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2
\]
\[
= -\frac{18}{16} + \frac{9}{4} + 2
\]
\[
= -\frac{9}{8} + \frac{9}{4} + 2
\]
Convert \(\frac{9}{4}\) and \(2\) to eighths:
\[
= -\frac{9}{8} + \frac{18}{8} + \frac{16}{8}
\]
\[
= -\frac{9}{8} + \frac{18 + 16}{8}
\]
\[
= -\frac{9}{8} + \frac{34}{8} = \frac{34 - 9}{8} = \frac{25}{8} \quad (\neq 0)
\]

### Testing \(x = -\frac{1}{2}\):
\[
-2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2
\]
\[
= -\frac{2}{4} + \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} + \frac{3}{2} + 2
\]
Convert \(2\) to halves:
\[
= -\frac{1}{2} + \frac{3}{2} + \frac{4}{2}
\]
\[
= -\frac{1}{2} + \frac{7}{2} = \frac{7 - 1}{2} = \frac{6}{2} = 3 \quad (\neq 0)
\]

### Testing \(x = \frac{1}{2}\):
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2
\]
\[
= -\frac{2}{4} - \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} - \frac{3}{2} + 2
\]
Convert \(2\) to halves:
\[
= -\frac{1}{2} - \frac{3}{2} + \frac{4}{2}
\]
\[
= -\frac{1 + 3 - 4}{2} = \frac{-4 + 4}{2} = \frac{0}{2} = 0 \quad (= 0)
\]

### Summary of Results:
- For \(x = -\frac{3}{4}\), the result is \(\frac{25}{8}\) (not a solution).
- For \(x = -\frac{1}{2}\), the result is \(3\) (not a solution).
- For \(x = \frac{1}{2}\), the result is \(0\) (this is a solution).

Thus, the value that is in the solution set is \(x = \frac{1}{2}\).