Find the Roots (Zeros) f(x)=x^2-6x-4

$f (x) = x^{2} - 6 x - 4$

Set $x^{2} - 6 x - 4$ equal to $0$ .

$x^{2} - 6 x - 4 = 0$

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 $a = 1$ , $b = - 6$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 4)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{6 \pm \sqrt{36 + 16}}{2 \cdot 1}$

Add $36$ and $16$ .

$x = \frac{6 \pm \sqrt{52}}{2 \cdot 1}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{6 \pm \sqrt{4 (13)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{13}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{13}}{2}$ .

$x = 3 \pm \sqrt{13}$

The final answer is the combination of both solutions.

$x = 3 + \sqrt{13}, 3 - \sqrt{13}$

$x = 3 \pm \sqrt{13}$

The result can be shown in multiple forms.

Exact Form:

$x = 3 \pm \sqrt{13}$

Decimal Form:

$x = 6.60555127 \dots, - 0.60555127 \dots$