Solve Using the Quadratic Formula x^2+4x-13=0

$x^{2} + 4 x - 13 = 0$

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 $a = 1$ , $b = 4$ , and $c = - 13$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 13)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 13}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 13$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 13}}{2 \cdot 1}$

Multiply $- 4$ by $- 13$ .

$x = \frac{- 4 \pm \sqrt{16 + 52}}{2 \cdot 1}$

Add $16$ and $52$ .

$x = \frac{- 4 \pm \sqrt{68}}{2 \cdot 1}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{- 4 \pm \sqrt{4 (17)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 17}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{17}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{17}}{2}$ .

$x = - 2 \pm \sqrt{17}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - 2 \pm \sqrt{17}$

Decimal Form:

$x = 2.12310562 \dots, - 6.12310562 \dots$