To determine whether -1 and \( \frac{7}{3} \) are solutions to the equation \( 3x^2 - 4x - 4 = 3 \), we first simplify the equation.
Subtract 3 from both sides:
\[ 3x^2 - 4x - 4 - 3 = 0 \] \[ 3x^2 - 4x - 7 = 0 \]
Now, let's check each value:
- Checking \( x = -1 \):
\[ 3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0 \]
The left side equals 0, so \( x = -1 \) is a solution.
- Checking \( x = \frac{7}{3} \):
Substituting \( x = \frac{7}{3} \):
\[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 \]
Calculating \( 3\left(\frac{7}{3}\right)^2 \):
\[ 3 \cdot \frac{49}{9} = \frac{147}{9} \]
Calculating \( -4\left(\frac{7}{3}\right) \):
\[ -\frac{28}{3} = -\frac{84}{9} \]
Now we put it all together:
\[ \frac{147}{9} - \frac{84}{9} - 7 = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0 \]
The left side also equals 0, so \( x = \frac{7}{3} \) is a solution.
Since both \( x = -1 \) and \( x = \frac{7}{3} \) satisfy the equation, the answer is:
D. Both x = -1 and x = 7/3 are solutions.