Question 5 (1 point) Solve. $$ 2 x^{2}-39=33 $$ a $$ \pm 3 $$ b no real-number solution c $$ \pm 6 $$ d $$ \pm \sqrt{6} $$ Question 6 (1 point) Solve: $$ -4(x-2)^{2}+3=-397_{\text {by square roots. }} $$ $$ \mathrm{x}= $$ $$ \qquad $$ and $$ \qquad $$ Blank 1: $$ \square $$ Blank 2: $$ \square $$

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**Question 5**

We start with the equation:
$$
2x^{2} - 39 = 33.
$$

1. Add $$39$$ to both sides:
$$
2x^{2} = 33 + 39 = 72.
$$

2. Divide both sides by $$2$$:
$$
x^{2} = \frac{72}{2} = 36.
$$

3. Take the square root of both sides:
$$
x = \pm \sqrt{36} = \pm 6.
$$

Thus, the solution is $$ \pm 6$$, which corresponds to option (c).

---

**Question 6**

We start with the equation:
$$
-4(x-2)^{2} + 3 = -397.
$$

1. Subtract $$3$$ from both sides:
$$
-4(x-2)^{2} = -397 - 3 = -400.
$$

2. Divide both sides by $$-4$$:
$$
(x-2)^{2} = \frac{-400}{-4} = 100.
$$

3. Take the square root of both sides:
$$
x-2 = \pm \sqrt{100} = \pm 10.
$$

4. Solve for $$x$$ by adding $$2$$ to both cases:
$$
x = 2 + 10 = 12 \quad \text{or} \quad x = 2 - 10 = -8.
$$

Thus, the solutions are:
- Blank 1: $$12$$
- Blank 2: $$-8$$
**Question 5** Solve: $$ 2x^{2} - 39 = 33 $$ - $$ \pm 6 $$ **Question 6** Solve: $$ -4(x-2)^{2} + 3 = -397 $$ - $$ x = 12 $$ and $$ x = -8 $$

Question 5 (1 point) Solve. \[ 2 x^{2}-39=33 \] a \( \pm 3 \) b no real-number solution c \( \pm 6 \) d \( \pm \sqrt{6} \) Question 6 (1 point) Solve: \( -4(x-2)^{2}+3=-397_{\text {by square roots. }} \) \( \mathrm{x}= \) \( \qquad \) and \( \qquad \) Blank 1: \( \square \) Blank 2: \( \square \)

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