Solve for k 3k^2-4k-1=0

$3 k^{2} - 4 k - 1 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 4$ , and $c = - 1$ into the quadratic formula and solve for $k$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (3 \cdot - 1)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$k = \frac{4 \pm \sqrt{16 - 4 \cdot 3 \cdot - 1}}{2 \cdot 3}$

Multiply $- 4$ by $3$ .

$k = \frac{4 \pm \sqrt{16 - 12 \cdot - 1}}{2 \cdot 3}$

Multiply $- 12$ by $- 1$ .

$k = \frac{4 \pm \sqrt{16 + 12}}{2 \cdot 3}$

Add $16$ and $12$ .

$k = \frac{4 \pm \sqrt{28}}{2 \cdot 3}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$k = \frac{4 \pm \sqrt{4 (7)}}{2 \cdot 3}$

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

$k = \frac{4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 3}$

Pull terms out from under the radical.

$k = \frac{4 \pm 2 \sqrt{7}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$k = \frac{4 \pm 2 \sqrt{7}}{6}$

Simplify $\frac{4 \pm 2 \sqrt{7}}{6}$ .

$k = \frac{2 \pm \sqrt{7}}{3}$

The final answer is the combination of both solutions.

$k = \frac{2 + \sqrt{7}}{3}, \frac{2 - \sqrt{7}}{3}$

The result can be shown in multiple forms.

Exact Form:

$k = \frac{2 + \sqrt{7}}{3}, \frac{2 - \sqrt{7}}{3}$

Decimal Form:

$k = 1.54858377 \dots, - 0.21525043 \dots$