To determine whether the values \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we first simplify the equation:
1. Start with the equation:
\[
3x^2 - 4x - 4 = 3
\]
2. Subtract 3 from both sides to set the equation to zero:
\[
3x^2 - 4x - 4 - 3 = 0
\]
\[
3x^2 - 4x - 7 = 0
\]
Now, we will test the given values one by one.
### Testing \( x = -1 \):
Substituting \( x = -1 \) into the left side of the equation:
\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7
\]
\[
= 3 + 4 - 7 = 0
\]
Since we get 0, \( x = -1 \) is a solution.
### Testing \( x = \frac{7}{3} \):
Substituting \( x = \frac{7}{3} \) into the left side of the equation:
\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7
\]
Calculating \( \left(\frac{7}{3}\right)^2 \):
\[
\left(\frac{7}{3}\right)^2 = \frac{49}{9}
\]
Then:
\[
3\left(\frac{49}{9}\right) - 4\left(\frac{7}{3}\right) - 7 = \frac{147}{9} - \frac{28}{3} - 7
\]
Converting \( \frac{28}{3} \) to a fraction with a denominator of 9:
\[
\frac{28}{3} = \frac{84}{9}
\]
And \( 7 = \frac{63}{9} \):
\[
\frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]
Since we also get 0 here, \( x = \frac{7}{3} \) is a solution.
### Conclusion:
Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the equation \( 3x^2 - 4x - 7 = 0 \).
Thus, the correct answer is:
**Both x = -1 and x = 7/3 are solutions.**