Use the quadratic formula to find both solutions to the quadratic equation given below.
$$3x^2-x+4=0$$
A. $x=\frac{1+\sqrt{48}}{6}$
B. $x=\frac{1+\sqrt{12}}{2}$
C. $x=\frac{1-\sqrt{-47}}{6}$
D. $x=\frac{1-\sqrt{12}}{2}$
E. $x=\frac{1-\sqrt{48}}{6}$
F. $x=\frac{1+\sqrt{-47}}{6}$

Step 1 ：Given the quadratic equation $3x^2-x+4=0$, we can identify the coefficients as $a=3$, $b=-1$, and $c=4$.

Step 2 ：We can use the quadratic formula to find the solutions for $x$, which is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Step 3 ：Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get $x=\frac{1\pm \sqrt{(-1{)}^2-4\ast 3\ast 4}}{2\ast 3}$.

Step 4 ：Simplifying the expression under the square root, we find that the discriminant (D) is -47, which is a negative number. This means the solutions will be complex numbers.

Step 5 ：The solutions can be written in the form $x=\frac{1\pm \sqrt{-47}}{6}$.

Step 6 ：Final Answer: The solutions to the quadratic equation are $\boxed{{\displaystyle x=\frac{1-\sqrt{-47}}{6}}}$ and $\boxed{{\displaystyle x=\frac{1+\sqrt{-47}}{6}}}$.