Solve Using the Quadratic Formula 6x^2+4x+11=0

$6 x^{2} + 4 x + 11 = 0$

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 $a = 6$ , $b = 4$ , and $c = 11$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 6 \cdot 11}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 11$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 4 \pm \sqrt{16 - 24 \cdot 11}}{2 \cdot 6}$

Multiply $- 24$ by $11$ .

$x = \frac{- 4 \pm \sqrt{16 - 264}}{2 \cdot 6}$

Subtract $264$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 248}}{2 \cdot 6}$

Rewrite $- 248$ as $- 1 (248)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 248}}{2 \cdot 6}$

Rewrite $\sqrt{- 1 (248)}$ as $\sqrt{- 1} \cdot \sqrt{248}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{248}}{2 \cdot 6}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 4 \pm i \cdot \sqrt{248}}{2 \cdot 6}$

Rewrite $248$ as $2^{2} \cdot 62$ .

Factor $4$ out of $248$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (62)}}{2 \cdot 6}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 62}}{2 \cdot 6}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{62})}{2 \cdot 6}$

Move $2$ to the left of $i$ .

$x = \frac{- 4 \pm 2 i \sqrt{62}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 4 \pm 2 i \sqrt{62}}{12}$

Simplify $\frac{- 4 \pm 2 i \sqrt{62}}{12}$ .

$x = \frac{- 2 \pm i \sqrt{62}}{6}$