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: \( \square \)

To tackle the first equation, let’s simplify it: \[ 2x^2 - 39 = 33 \] Add 39 to both sides: \[ 2x^2 = 72 \] Divide by 2: \[ x^2 = 36 \] Now, taking the square root of both sides gives us \( x = \pm 6 \). So, the correct option is: **c \( \pm 6 \)** Now, on to the second question. Start with: \[ -4(x-2)^{2} + 3 = -397 \] First, isolate the squared term: \[ -4(x-2)^{2} = -400 \] Divide by -4: \[ (x-2)^{2} = 100 \] Now, take the square root: \[ x - 2 = \pm 10 \] So, \( x = 12 \) or \( x = -8 \). In the blanks, you'd have: Blank 1: \( 12 \) Blank 2: \( -8 \)