Solve Using the Quadratic Formula 6x^2+2x-1=0

$6 x^{2} + 2 x - 1 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 6$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 6 \cdot - 1$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 2 \pm \sqrt{4 - 24 \cdot - 1}}{2 \cdot 6}$

Multiply $- 24$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 24}}{2 \cdot 6}$

Add $4$ and $24$ .

$x = \frac{- 2 \pm \sqrt{28}}{2 \cdot 6}$

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

Factor $4$ out of $28$ .

$x = \frac{- 2 \pm \sqrt{4 (7)}}{2 \cdot 6}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 6}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{7}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 2 \pm 2 \sqrt{7}}{12}$

Simplify $\frac{- 2 \pm 2 \sqrt{7}}{12}$ .

$x = \frac{- 1 \pm \sqrt{7}}{6}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{7}}{6}$

Decimal Form:

$x = 0.27429188 \dots, - 0.60762521 \dots$