To determine whether each of the given values is a solution to the quadratic equation \(-2x^2 - 3x + 2 = 0\), we will substitute each value into the equation and evaluate it.
### Step 1: Substitute each value into the equation
1. **For \(x = -\frac{3}{4}\)**:
\[
-2\left(-\frac{3}{4}\right)^2 - 3\left(-\frac{3}{4}\right) + 2
\]
\[
= -2\left(\frac{9}{16}\right) + \frac{9}{4} + 2
\]
\[
= -\frac{18}{16} + \frac{36}{16} + \frac{32}{16}
\]
\[
= -\frac{18}{16} + \frac{68}{16}
\]
\[
= \frac{50}{16} \neq 0
\]
2. **For \(x = -\frac{1}{2}\)**:
\[
-2\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) + \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} + \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} + \frac{3}{2} + \frac{4}{2}
\]
\[
= \frac{6}{2} - \frac{1}{2}
\]
\[
= \frac{5}{2} \neq 0
\]
3. **For \(x = \frac{1}{2}\)**:
\[
-2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 2
\]
\[
= -2\left(\frac{1}{4}\right) - \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} - \frac{3}{2} + 2
\]
\[
= -\frac{1}{2} - \frac{3}{2} + \frac{4}{2}
\]
\[
= \frac{4}{2} - \frac{4}{2} = 0
\]
### Step 2: Fill the table and identify solutions
| x | Substituted value | Evaluate | True Statement? |
|--------|-------------------------|-----------------------|----------------------|
| -3/4 | -2(-3/4)² - 3(-3/4) + 2 | 50/16 \(\neq 0\) | False |
| -1/2 | -2(-1/2)² - 3(-1/2) + 2 | 5/2 \(\neq 0\) | False |
| 1/2 | -2(1/2)² - 3(1/2) + 2 | 0 | True |
### Conclusion
Only \( x = \frac{1}{2} \) is in the solution set.
So, the answer is:
**Only x= ( \(\frac{1}{2}\) ) is in the solution set.**