This is a quadratic equation. There's a shortcut (two shortcuts really) called the Shakespeare Quadratic Formula (2b or -2b) applicable when the middle term is even:
[tex]x^2 - 2bx + c \textrm{ has zeros } x= b\pm \sqrt{b^2-c}[/tex]
[tex]ax^2 - 2bx + c \textrm{ has zeros } x= \dfrac 1 a( \left b\pm \sqrt{b^2-ac} \right)[/tex]
OK, our problem is
[tex]9x^2 - 6x -7 = 0[/tex]
[tex]x = \frac 1 9 (3 \pm \sqrt{9-(9)(-7)}) = \frac 1 9(3 \pm \sqrt{72})= \frac 1 9(3 \pm 3 \sqrt{8}) = \frac 1 3(1 \pm \sqrt 8)[/tex]
Typically we'd write [tex]\sqrt 8 = 2 \sqrt 2[/tex] but the answers here don't so we won't either.
Answer: C and D
