Solve Using the Quadratic Formula x^2+6x-11=0

$x^{2} + 6 x - 11 = 0$

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 $a = 1$ , $b = 6$ , and $c = - 11$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot - 11)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 11}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 11$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 11}}{2 \cdot 1}$

Multiply $- 4$ by $- 11$ .

$x = \frac{- 6 \pm \sqrt{36 + 44}}{2 \cdot 1}$

Add $36$ and $44$ .

$x = \frac{- 6 \pm \sqrt{80}}{2 \cdot 1}$

Rewrite $80$ as $4^{2} \cdot 5$ .

Factor $16$ out of $80$ .

$x = \frac{- 6 \pm \sqrt{16 (5)}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm \sqrt{4^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm 4 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 6 \pm 4 \sqrt{5}}{2}$

Simplify $\frac{- 6 \pm 4 \sqrt{5}}{2}$ .

$x = - 3 \pm 2 \sqrt{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - 3 \pm 2 \sqrt{5}$

Decimal Form:

$x = 1.47213595 \dots, - 7.47213595 \dots$