Determine whether the values −1 and 7/3 are solutions to the quadratic equation 3x2−4x−4=3 .(1 point)

To determine if the values \(-1\) and \(\frac{7}{3}\) are solutions to the equation \(3x^2 - 4x - 4 = 3\), we first need to 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 check if each value satisfies this equation.

### Check \(x = -1\):

Substituting \(-1\) into the equation:

\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7
\]
\[
= 3 + 4 - 7 = 0
\]

Since the left side equals 0, \(-1\) is a solution.

### Check \(x = \frac{7}{3}\):

Substituting \(\frac{7}{3}\) into the equation:

\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 = 3\left(\frac{49}{9}\right) - \frac{28}{3} - 7
\]
\[
= \frac{147}{9} - \frac{84}{9} - \frac{63}{9}
\]
\[
= \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]

Since the left side also equals 0, \(\frac{7}{3}\) is also a solution.

### Conclusion:
Both values \(-1\) and \(\frac{7}{3}\) are solutions to the equation \(3x^2 - 4x - 7 = 0\).