Solve Using the Quadratic Formula 2x^2+4x-6=0

$2 x^{2} + 4 x - 6 = 0$

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 $a = 2$ , $b = 4$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 2 \cdot - 6}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 6$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 8 \cdot - 6}}{2 \cdot 2}$

Multiply $- 8$ by $- 6$ .

$x = \frac{- 4 \pm \sqrt{16 + 48}}{2 \cdot 2}$

Add $16$ and $48$ .

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

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 4 \pm \sqrt{8^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 4 \pm 8}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 4 \pm 8}{4}$

Simplify $\frac{- 4 \pm 8}{4}$ .

$x = - 1 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 3$