Use the quadratic formula to find both solutions to the quadratic equation given below.
\[
3 x^{2}-x+4=0
\]
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 \(x=\frac{1\pm\sqrt{(-1)^{2}-4*3*4}}{2*3}\).

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 \(\boxed{x=\frac{1-\sqrt{-47}}{6}}\) and \(\boxed{x=\frac{1+\sqrt{-47}}{6}}\).