Find All Complex Solutions 6x^2-3x+2=0

$6 x^{2} - 3 x + 2 = 0$

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 $a = 6$ , $b = - 3$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (6 \cdot 2)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 6 \cdot 2}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 2$ .

Multiply $- 4$ by $6$ .

$x = \frac{3 \pm \sqrt{9 - 24 \cdot 2}}{2 \cdot 6}$

Multiply $- 24$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 48}}{2 \cdot 6}$

Subtract $48$ from $9$ .

$x = \frac{3 \pm \sqrt{- 39}}{2 \cdot 6}$

Rewrite $- 39$ as $- 1 (39)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 39}}{2 \cdot 6}$

Rewrite $\sqrt{- 1 (39)}$ as $\sqrt{- 1} \cdot \sqrt{39}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{39}}{2 \cdot 6}$

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

$x = \frac{3 \pm i \sqrt{39}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{3 \pm i \sqrt{39}}{12}$