How do you solve $4{x^2} - 4x - 1 = 0$ ?

Hint: The given problem requires us to solve a quadratic equation. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease.

Complete step by step answer:
In the given question, we are required to solve the equation $4{x^2} - 4x - 1 = 0$ with the help of a quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.
Comparing with standard quadratic equation $a{x^2} + bx + c = 0$
Here,$a = 4$, $b = - 4$ and$c = - 1$.
Now, using the quadratic formula, we get the roots of the equation as:
$x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values of a, b, and c in the quadratic formula, we get,
$x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{( - 4)}^2} - 4 \times 4 \times \left( { - 1} \right)} }}{{2 \times 4}}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 16} }}{{2 \times 4}}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {32} }}{8}$
Simplifying the square root as we know that $32 = 16 \times 2$.
$ \Rightarrow x = \dfrac{{4 \pm 4\sqrt 2 }}{8}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow x = \dfrac{1}{2} \pm \dfrac{{\sqrt 2 }}{2}$
Simplifying the roots further,
$ \Rightarrow x = \dfrac{1}{2} \pm \dfrac{1}{{\sqrt 2 }}$
So, $x = \dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }}$ and $x = \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }}$ are the roots of the equation $4{x^2} - 4x - 1 = 0$.
Therefore, the roots of the equation $4{x^2} - 4x - 1 = 0$ are: $x = \dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }}$ and $x = \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }}$.

Note:
Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic formula is the easiest and most efficient formula to calculate the roots of an equation. Quadratic equations can also be solved by splitting the middle term and completing the square method. Quadratic equations may also be solved by a hit and trial method if the roots of the equation are easy to find.