Find All Complex Solutions 2x^2-5x+11=0

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

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 $a = 2$ , $b = - 5$ , and $c = 11$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot 11$ .

Multiply $- 4$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 8 \cdot 11}}{2 \cdot 2}$

Multiply $- 8$ by $11$ .

$x = \frac{5 \pm \sqrt{25 - 88}}{2 \cdot 2}$

Subtract $88$ from $25$ .

$x = \frac{5 \pm \sqrt{- 63}}{2 \cdot 2}$

Rewrite $- 63$ as $- 1 (63)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 63}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (63)}$ as $\sqrt{- 1} \cdot \sqrt{63}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{63}}{2 \cdot 2}$

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

$x = \frac{5 \pm i \cdot \sqrt{63}}{2 \cdot 2}$

Rewrite $63$ as $3^{2} \cdot 7$ .

Factor $9$ out of $63$ .

$x = \frac{5 \pm i \cdot \sqrt{9 (7)}}{2 \cdot 2}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{5 \pm i \cdot \sqrt{3^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{5 \pm i \cdot (3 \sqrt{7})}{2 \cdot 2}$

Move $3$ to the left of $i$ .

$x = \frac{5 \pm 3 i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5 \pm 3 i \sqrt{7}}{4}$