Answer:
option D
Step-by-step explanation:
[tex]9x^2 + 6x - 17 = 0 , \ where \ a = 9 \ , \ b = 6 \ , c = - 17[/tex]
[tex]x = \frac{ - b \ \pm \sqrt{b^2 - 4ac}}{2a} \\\\x = \frac{-6 \pm \sqrt{36 - (4 \times 9 \times -17)}}{2 \times 9}\\\\x = \frac{-6 \pm \sqrt{ 36 + 612}}{18}\\\\x = \frac{-6 \pm \sqrt{ 648}}{18}\\\\x = \frac{-6 \pm \sqrt{ 2^3 \times 3^4}}{18}\\\\x = \frac{-6 \pm \sqrt{2 \times 2^2 \times 3^ 4 }}{18}\\\\x = \frac{-6 \pm ( 2 \times 9 )\sqrt{ 2}}{18}\\\\x = \frac{-6 \pm 18\sqrt{2}}{18}\\\\x = \frac{-1 \pm 3 \sqrt{ 2}}{3}[/tex]
