To determine whether the values \( -1 \) and \( \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we first rewrite the equation in standard form by moving all terms to one side:
\[ 3x^2 - 4x - 4 - 3 = 0 \]
This simplifies to:
\[ 3x^2 - 4x - 7 = 0 \]
Now, we will evaluate both \( x = -1 \) and \( x = \frac{7}{3} \) in this equation.
Check \( x = -1 \):
\[ 3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0 \]
So, \( x = -1 \) is a solution.
Check \( x = \frac{7}{3} \):
First, find \( \left( \frac{7}{3} \right)^2 \):
\[ \left( \frac{7}{3} \right)^2 = \frac{49}{9} \]
Now, substitute \( x = \frac{7}{3} \) into the equation:
\[ 3\left( \frac{7}{3} \right)^2 - 4\left( \frac{7}{3} \right) - 7 = 3\left( \frac{49}{9} \right) - \frac{28}{3} - 7 \]
Calculating \( 3\left( \frac{49}{9} \right) \):
\[ 3\left( \frac{49}{9} \right) = \frac{147}{9} \]
Next, convert \( -\frac{28}{3} \) and \( -7 \) to have a common denominator of 9:
\[ -\frac{28}{3} = -\frac{84}{9} \] \[ -7 = -\frac{63}{9} \]
Now substitute:
\[ \frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0 \]
So, \( x = \frac{7}{3} \) is also a solution.
Conclusion
Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the equation. Therefore, the correct response is:
Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions.