Answer
Solution set = $\displaystyle \{\frac{3-\sqrt{57}}{6},\frac{3+\sqrt{57}}{6}\}.$
Work Step by Step
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$3x^{2}-3x-4=0 \quad\rightarrow \left\{\begin{array}{l}
a=3\\
b=-3\\
c=-4
\end{array}\right.$
$x=\displaystyle \frac{-(-3)\pm\sqrt{(-3)^{2}-4(3)(-4)}}{2(3)}=\frac{3\pm\sqrt{9+48}}{6}=\frac{3\pm\sqrt{57}}{6}$
$x= \displaystyle \frac{3-\sqrt{57}}{6}\qquad$ or$ \displaystyle \qquad x=\frac{3+\sqrt{57}}{6}$
Solution set = $\displaystyle \{\frac{3-\sqrt{57}}{6},\frac{3+\sqrt{57}}{6}\}.$
