Solve Using the Quadratic Formula 2x^2+16x+28=0

$2 x^{2} + 16 x + 28 = 0$

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 $a = 2$ , $b = 16$ , and $c = 28$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (2 \cdot 28)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 2 \cdot 28}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 28$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 8 \cdot 28}}{2 \cdot 2}$

Multiply $- 8$ by $28$ .

$x = \frac{- 16 \pm \sqrt{256 - 224}}{2 \cdot 2}$

Subtract $224$ from $256$ .

$x = \frac{- 16 \pm \sqrt{32}}{2 \cdot 2}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{- 16 \pm \sqrt{16 (2)}}{2 \cdot 2}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $2$ .

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

Simplify $\frac{- 16 \pm 4 \sqrt{2}}{4}$ .

$x = - 4 \pm \sqrt{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - 4 \pm \sqrt{2}$

Decimal Form:

$x = - 2.58578643 \dots, - 5.41421356 \dots$