Find the Roots (Zeros) f(x)=x^2-2x-4

$f (x) = x^{2} - 2 x - 4$

Step 1

Set $x^{2} - 2 x - 4$ equal to $0$ .

$x^{2} - 2 x - 4 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 4)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 4$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

The final answer is the combination of both solutions.

$x = 1 + \sqrt{5}, 1 - \sqrt{5}$

$x = 1 \pm \sqrt{5}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 1 \pm \sqrt{5}$

Decimal Form:

$x = 3.23606797 \dots, - 1.23606797 \dots$

Step 4