Solve Using the Quadratic Formula x^2+12x+28=0

$x^{2} + 12 x + 28 = 0$

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 $a = 1$ , $b = 12$ , and $c = 28$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 28)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 28}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 28$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 28}}{2 \cdot 1}$

Multiply $- 4$ by $28$ .

$x = \frac{- 12 \pm \sqrt{144 - 112}}{2 \cdot 1}$

Subtract $112$ from $144$ .

$x = \frac{- 12 \pm \sqrt{32}}{2 \cdot 1}$

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

Factor $16$ out of $32$ .

$x = \frac{- 12 \pm \sqrt{16 (2)}}{2 \cdot 1}$

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

$x = \frac{- 12 \pm \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 12 \pm 4 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

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

Simplify $\frac{- 12 \pm 4 \sqrt{2}}{2}$ .

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 3.17157287 \dots, - 8.82842712 \dots$