fully Factorise $$ 2 x^{2}-4 x-1 $$

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Solve the quadratic equation by following steps:
- step0: Solve using the quadratic formula:
  $$2x^{2}-4x-1=0$$
- step1: Solve using the quadratic formula:
  $$x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$
- step2: Simplify the expression:
  $$x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 2\left(-1\right)}}{4}$$
- step3: Simplify the expression:
  $$x=\frac{4\pm \sqrt{24}}{4}$$
- step4: Simplify the expression:
  $$x=\frac{4\pm 2\sqrt{6}}{4}$$
- step5: Separate into possible cases:
  $$\begin{align}&x=\frac{4+2\sqrt{6}}{4}\&x=\frac{4-2\sqrt{6}}{4}\end{align}$$
- step6: Simplify the expression:
  $$\begin{align}&x=\frac{2+\sqrt{6}}{2}\&x=\frac{4-2\sqrt{6}}{4}\end{align}$$
- step7: Simplify the expression:
  $$\begin{align}&x=\frac{2+\sqrt{6}}{2}\&x=\frac{2-\sqrt{6}}{2}\end{align}$$
- step8: Rewrite:
  $$x_{1}=\frac{2-\sqrt{6}}{2},x_{2}=\frac{2+\sqrt{6}}{2}$$
To fully factor the expression $$ 2x^{2}-4x-1 $$, we first need to find its roots by solving the equation $$ 2x^{2}-4x-1=0 $$.

### Step 1: Solve the quadratic equation

Using the quadratic formula, we find the roots:

$$
x_{1} = \frac{2 - \sqrt{6}}{2}, \quad x_{2} = \frac{2 + \sqrt{6}}{2}
$$

### Step 2: Rewrite the quadratic expression

The roots $$ x_{1} $$ and $$ x_{2} $$ can be used to express the quadratic in its factored form. The general form of a quadratic can be expressed as:

$$
a(x - x_{1})(x - x_{2})
$$

where $$ a $$ is the leading coefficient (which is 2 in this case).

### Step 3: Substitute the roots into the factored form

Substituting the roots into the factored form gives us:

$$
2\left(x - \frac{2 - \sqrt{6}}{2}\right)\left(x - \frac{2 + \sqrt{6}}{2}\right)
$$

### Step 4: Simplify the factors

This can be simplified further:

$$
2\left(x - 1 + \frac{\sqrt{6}}{2}\right)\left(x - 1 - \frac{\sqrt{6}}{2}\right)
$$

### Final Factored Form

Thus, the fully factored form of $$ 2x^{2}-4x-1 $$ is:

$$
2\left(x - 1 + \frac{\sqrt{6}}{2}\right)\left(x - 1 - \frac{\sqrt{6}}{2}\right)
$$

This represents the complete factorization of the quadratic expression.
The fully factored form of $$ 2x^{2}-4x-1 $$ is $$ 2\left(x - 1 + \frac{\sqrt{6}}{2}\right)\left(x - 1 - \frac{\sqrt{6}}{2}\right) $$.

fully Factorise \( 2 x^{2}-4 x-1 \)

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