ANSWER
[tex]B)\text{ }x=\frac{1\pm2\sqrt[]{3}}{3}[/tex]
EXPLANATION
To solve this, first, we have to subtract 11 from both sides of the equation,
[tex]\begin{gathered} 9x^2-6x-11=11-11 \\ \\ 9x^2-6x-11=0 \end{gathered}[/tex]
Now, the equation is in the form,
[tex]ax^2+bx+c=0[/tex]
And we can apply the quadratic formula,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
In this case, a = 9, b = -6, and c = -11,
[tex]x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot9\cdot(-11)}}{2\cdot9}=\frac{6\pm\sqrt[]{36+396}}{18}=\frac{6\pm\sqrt[]{432}}{18}=\frac{6\pm12\sqrt[]{3}}{18}[/tex]
We can simplify this by factoring out 6 in both the numerator and denominator,
[tex]x=\frac{6(\frac{6}{6}\pm\frac{12}{6}\sqrt[]{3})}{6(\frac{18}{6})}=\frac{1\pm2\sqrt[]{3}}{3}[/tex]
Hence, the answer is option B.
