## Precalculus (6th Edition)

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