Let's tackle each problem one by one.

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Problem 1:

Determine whether the values $x = -1$ and $x = \frac{7}{3}$ are solutions to the quadratic equation $3x^2 - 4x - 4 = 3$.

1. Start by rearranging the equation:
   $3x^2 - 4x - 4 = 3 \quad \Rightarrow \quad 3x^2 - 4x - 7 = 0$

2. Substitute $x = -1$ into the equation: $3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 0$ This is true, so $x = -1$ is a solution.

3. Substitute $x = \frac{7}{3}$: $3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 = 3 \times \frac{49}{9} - \frac{28}{3} - 7 = \frac{147}{9} - \frac{84}{9} - \frac{63}{9} = 0$ This is also true, so $x = \frac{7}{3}$ is a solution.

Thus, both $x = -1$ and $x = \frac{7}{3}$ are solutions.

Answer: C. Both $x = -1$ and $x = \frac{7}{3}$ are solutions.

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Problem 2:

We are given the quadratic function for the height of a firework:
$h = -\frac{25}{36}x^2 + \frac{50}{3}x$ and need to find which value of $x$ makes $h = 0$ (i.e., when the firework reaches the ground).

Solve for $x$ in: $0 = -\frac{25}{36}x^2 + \frac{50}{3}x$ Factor out $x$: $x\left(-\frac{25}{36}x + \frac{50}{3}\right) = 0$ Set each factor to 0:

1. $x = 0$, which is the starting point.
2. Solve $-\frac{25}{36}x + \frac{50}{3} = 0$: $\frac{25}{36}x = \frac{50}{3} \quad \Rightarrow \quad x = \frac{50}{3} \times \frac{36}{25} = 24$

Thus, the firework reaches the ground when $x = 24$.

Answer: D. 24 feet.

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Problem 3:

Determine whether $x = -3$ is a solution to the equation $-3x^2 - 9x = 0$.

Substitute $x = -3$ into the equation: $-3(-3)^2 - 9(-3) = -3(9) + 27 = -27 + 27 = 0$ This is true, so $x = -3$ is a solution.

Answer: C. $x = -3$ is a solution because substituting it back into the equation results in $0 = 0$.

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Problem 4:

Solve for $x$ in the equation $3x^2 = 21$.

1. Divide both sides by 3: $x^2 = 7$
2. Take the square root of both sides: $x = \pm \sqrt{7}$

Thus, there are two solutions: $x = \sqrt{7}$ and $x = -\sqrt{7}$.

Answer: A. Two.

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Problem 5:

Solve for $x$ in the equation $2(x - 5)^2 + 2 = 20$.

1. Subtract 2 from both sides: $2(x - 5)^2 = 18$
2. Divide both sides by 2: $(x - 5)^2 = 9$
3. Take the square root of both sides: $x - 5 = \pm 3$
4. Solve for $x$: $x = 5 + 3 = 8 \quad \text{or} \quad x = 5 - 3 = 2$

Thus, there are two solutions: $x = 8$ and $x = 2$.

Answer: D. Two.

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Would you like further details on any of the solutions or clarifications?

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Here are 5 follow-up questions to deepen your understanding:

1. How do you determine whether a value is a solution to a quadratic equation?
2. What are the methods to solve a quadratic equation like $3x^2 - 4x - 7 = 0$?
3. Why do quadratic equations often have two solutions?
4. How can you solve a quadratic equation by factoring?
5. What is the significance of the discriminant in determining the number of solutions to a quadratic equation?

Tip: When solving quadratic equations, always check for multiple solutions by considering both the positive and negative square roots.