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

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

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 $a = 2$ , $b = 6$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 8 \cdot - 3}}{2 \cdot 2}$

Multiply $- 8$ by $- 3$ .

$x = \frac{- 6 \pm \sqrt{36 + 24}}{2 \cdot 2}$

Add $36$ and $24$ .

$x = \frac{- 6 \pm \sqrt{60}}{2 \cdot 2}$

Rewrite $60$ as $2^{2} \cdot 15$ .

Factor $4$ out of $60$ .

$x = \frac{- 6 \pm \sqrt{4 (15)}}{2 \cdot 2}$

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

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{15}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 6 \pm 2 \sqrt{15}}{4}$

Simplify $\frac{- 6 \pm 2 \sqrt{15}}{4}$ .

$x = \frac{- 3 \pm \sqrt{15}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 3 \pm \sqrt{15}}{2}$

Decimal Form:

$x = 0.43649167 \dots, - 3.43649167 \dots$