Solve Using the Quadratic Formula 14x^2+3x-2=0

$14 x^{2} + 3 x - 2 = 0$

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 $a = 14$ , $b = 3$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (14 \cdot - 2)}}{2 \cdot 14}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 14 \cdot - 2}}{2 \cdot 14}$

Multiply $- 4 \cdot 14 \cdot - 2$ .

Multiply $- 4$ by $14$ .

$x = \frac{- 3 \pm \sqrt{9 - 56 \cdot - 2}}{2 \cdot 14}$

Multiply $- 56$ by $- 2$ .

$x = \frac{- 3 \pm \sqrt{9 + 112}}{2 \cdot 14}$

Add $9$ and $112$ .

$x = \frac{- 3 \pm \sqrt{121}}{2 \cdot 14}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{- 3 \pm \sqrt{11^{2}}}{2 \cdot 14}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 3 \pm 11}{2 \cdot 14}$

Multiply $2$ by $14$ .

$x = \frac{- 3 \pm 11}{28}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{2}{7}, - \frac{1}{2}$