Solve Using the Quadratic Formula 3x^2-x+4=0

$3 x^{2} - x + 4 = 0$

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 $a = 3$ , $b = - 1$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

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

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

Multiply $- 4$ by $3$ .

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

Multiply $- 12$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 48}}{2 \cdot 3}$

Subtract $48$ from $1$ .

$x = \frac{1 \pm \sqrt{- 47}}{2 \cdot 3}$

Rewrite $- 47$ as $- 1 (47)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 47}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (47)}$ as $\sqrt{- 1} \cdot \sqrt{47}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{47}}{2 \cdot 3}$

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

$x = \frac{1 \pm i \sqrt{47}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{1 \pm i \sqrt{47}}{6}$