Find the Roots (Zeros) x^2+5x+11

$x^{2} + 5 x + 11$

Step 1

Set $x^{2} + 5 x + 11$ equal to $0$ .

$x^{2} + 5 x + 11 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 5$ , and $c = 11$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot 11)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot 11}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 11}}{2 \cdot 1}$

Multiply $- 4$ by $11$ .

$x = \frac{- 5 \pm \sqrt{25 - 44}}{2 \cdot 1}$

Subtract $44$ from $25$ .

$x = \frac{- 5 \pm \sqrt{- 19}}{2 \cdot 1}$

Rewrite $- 19$ as $- 1 (19)$ .

$x = \frac{- 5 \pm \sqrt{- 1 \cdot 19}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (19)}$ as $\sqrt{- 1} \cdot \sqrt{19}$ .

$x = \frac{- 5 \pm \sqrt{- 1} \cdot \sqrt{19}}{2 \cdot 1}$

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

$x = \frac{- 5 \pm i \sqrt{19}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm i \sqrt{19}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{5 - i \sqrt{19}}{2}, - \frac{5 + i \sqrt{19}}{2}$

$x = \frac{- 5 \pm i \sqrt{19}}{2}$

Step 3