Answer:
D is the answer.
[tex]16x {}^{2} - 81 = 0 \\ (4x - 9) \times (4x + 9) = 0 \\ [/tex]
[tex]4x - 9 = 0 \\ 4x = 9 \\ x = \frac{9}{4} \\ or \\ 4x + 9 = 0 \\ 4x = - 9 \\ x = - \frac{9}{4} [/tex]
