Solve Using the Quadratic Formula 3x^2+2x-4=0

$3 x^{2} + 2 x - 4 = 0$

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 $a = 3$ , $b = 2$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot - 4}}{2 \cdot 3}$

Multiply $- 12$ by $- 4$ .

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot 3}$

Add $4$ and $48$ .

$x = \frac{- 2 \pm \sqrt{52}}{2 \cdot 3}$

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

Factor $4$ out of $52$ .

$x = \frac{- 2 \pm \sqrt{4 (13)}}{2 \cdot 3}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{13}}{2 \cdot 3}$

Multiply $2$ by $3$ .

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

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

$x = \frac{- 1 \pm \sqrt{13}}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{13}}{3}$

Decimal Form:

$x = 0.86851709 \dots, - 1.53518375 \dots$