Solve Using the Quadratic Formula 6x^2-2x-3=0

$6 x^{2} - 2 x - 3 = 0$

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 $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (6 \cdot - 3)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 6 \cdot - 3}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 3$ .

Multiply $- 4$ by $6$ .

$x = \frac{2 \pm \sqrt{4 - 24 \cdot - 3}}{2 \cdot 6}$

Multiply $- 24$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 72}}{2 \cdot 6}$

Add $4$ and $72$ .

$x = \frac{2 \pm \sqrt{76}}{2 \cdot 6}$

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

Factor $4$ out of $76$ .

$x = \frac{2 \pm \sqrt{4 (19)}}{2 \cdot 6}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 19}}{2 \cdot 6}$

Pull terms out from under the radical.

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

Multiply $2$ by $6$ .

$x = \frac{2 \pm 2 \sqrt{19}}{12}$

Simplify $\frac{2 \pm 2 \sqrt{19}}{12}$ .

$x = \frac{1 \pm \sqrt{19}}{6}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{19}}{6}$

Decimal Form:

$x = 0.89314982 \dots, - 0.55981649 \dots$