Solve Using the Quadratic Formula 4x^2-3x+6=0

$4 x^{2} - 3 x + 6 = 0$

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 $a = 4$ , $b = - 3$ , and $c = 6$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $4$ .

$x = \frac{3 \pm \sqrt{9 - 16 \cdot 6}}{2 \cdot 4}$

Multiply $- 16$ by $6$ .

$x = \frac{3 \pm \sqrt{9 - 96}}{2 \cdot 4}$

Subtract $96$ from $9$ .

$x = \frac{3 \pm \sqrt{- 87}}{2 \cdot 4}$

Rewrite $- 87$ as $- 1 (87)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 87}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (87)}$ as $\sqrt{- 1} \cdot \sqrt{87}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{87}}{2 \cdot 4}$

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

$x = \frac{3 \pm i \sqrt{87}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{3 \pm i \sqrt{87}}{8}$