How do you solve \[3{{x}^{2}}-x-1=0\] using the quadratic formula?

Hint: Assume the coefficient of \[{{x}^{2}}\] as ‘a’, coefficient of x as ‘b’ and constant term as ‘c’. Now, apply the discriminant formula given as: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to find the two values of x and get the answer.

Complete step by step answer:
Here, we have been provided with the quadratic equation \[3{{x}^{2}}-x-1=0\] and we are asked to find the values of x using the quadratic formula. But first, we need to know about the quadratic formula.
Now, in elementary algebra, the quadratic formula is an expression that provides the solution of a quadratic equation. There are some other methods also by which we can solve a quadratic equation like factoring, graphing, completing the square method, etc. But sometimes it is difficult to factor as we are not able to split the middle term, in such cases we use the quadratic formula.
Let us come to the question. We have the quadratic equation: - \[3{{x}^{2}}-x-1\], so assuming the coefficient of \[{{x}^{2}}\] as a, coefficient of x as b and the constant term as c, we have,
\[\Rightarrow \] a = 3, b = -1, c = -1
According to the quadratic formula or discriminant formula the values of x is given as: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. So, substituting the values of a, b and c we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times \left( 3 \right)\left( -1 \right)}}{2\times 3} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{1+12}}{6} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{13}}{6} \\
\end{align}\]
Hence, the two values of x are \[\dfrac{1+\sqrt{13}}{6}\] and \[\dfrac{1-\sqrt{13}}{6}\].

Note:
One may note that it was not possible for us to use the middle term split method to determine the values of x. This is because it is difficult to think of the roots like \[\dfrac{1+\sqrt{13}}{6}\] and \[\dfrac{1-\sqrt{13}}{6}\]. Also, we cannot apply the graphical method to get the answer because in the end, we will need the quadratic formula to get the points, i.e., the values of x. Here, we can apply to complete the square method to get our answer. Note that the discriminant formula is derived from completing the square method.