Solve Using the Quadratic Formula 3x^2+8x-1=0

$3 x^{2} + 8 x - 1 = 0$

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 $a = 3$ , $b = 8$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (3 \cdot - 1)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot - 1}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 1$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 12 \cdot - 1}}{2 \cdot 3}$

Multiply $- 12$ by $- 1$ .

$x = \frac{- 8 \pm \sqrt{64 + 12}}{2 \cdot 3}$

Add $64$ and $12$ .

$x = \frac{- 8 \pm \sqrt{76}}{2 \cdot 3}$

Rewrite $76$ as $2^{2} \cdot 19$ .

Factor $4$ out of $76$ .

$x = \frac{- 8 \pm \sqrt{4 (19)}}{2 \cdot 3}$

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

$x = \frac{- 8 \pm \sqrt{2^{2} \cdot 19}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm 2 \sqrt{19}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 8 \pm 2 \sqrt{19}}{6}$

Simplify $\frac{- 8 \pm 2 \sqrt{19}}{6}$ .

$x = \frac{- 4 \pm \sqrt{19}}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 4 \pm \sqrt{19}}{3}$

Decimal Form:

$x = 0.11963298 \dots, - 2.78629964 \dots$