Find All Complex Solutions 3x^2-2x+1=0

$3 x^{2} - 2 x + 1 = 0$

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 $a = 3$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (3 \cdot 1)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 1$ .

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12 \cdot 1}}{2 \cdot 3}$

Multiply $- 12$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 3}$

Subtract $12$ from $4$ .

$x = \frac{2 \pm \sqrt{- 8}}{2 \cdot 3}$

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 3}$

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

$x = \frac{2 \pm i \cdot \sqrt{8}}{2 \cdot 3}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 3}$

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $3$ .

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

Simplify $\frac{2 \pm 2 i \sqrt{2}}{6}$ .

$x = \frac{1 \pm i \sqrt{2}}{3}$