#### Answer

Option (A)
$x=\left\{-\frac{1}{3}, 6\right\}$

#### Work Step by Step

The equation in Option (A) has $a=3, b=-17, c=-6$.
Use the quadratic formula $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2}$ using the values of a, b, and c above to obtain:
$x = \dfrac{-(-17) \pm \sqrt{(-17)^2-4(3)(-6)}}{2(3)}
\\x = \dfrac{17 \pm \sqrt{289-(-72)}}{6}
\\x=\dfrac{17\pm\sqrt{289+72}}{6}
\\x=\dfrac{17\pm\sqrt{361}}{6}
\\x=\dfrac{17\pm 19}{6}
\\x_1=\dfrac{17+19}{6} = \dfrac{36}{6}=6
\\x_2=\dfrac{17-19}{6} = \dfrac{-2}{6} = -\dfrac{1}{3}$
Thus, the solutions to the given equation are:
$x=\left\{-\frac{1}{3}, 6\right\}$