Solve Using the Quadratic Formula 14x^2+9x+1=0

$14 x^{2} + 9 x + 1 = 0$

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 $a = 14$ , $b = 9$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (14 \cdot 1)}}{2 \cdot 14}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 14 \cdot 1}}{2 \cdot 14}$

Multiply $- 4 \cdot 14 \cdot 1$ .

Multiply $- 4$ by $14$ .

$x = \frac{- 9 \pm \sqrt{81 - 56 \cdot 1}}{2 \cdot 14}$

Multiply $- 56$ by $1$ .

$x = \frac{- 9 \pm \sqrt{81 - 56}}{2 \cdot 14}$

Subtract $56$ from $81$ .

$x = \frac{- 9 \pm \sqrt{25}}{2 \cdot 14}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 9 \pm \sqrt{5^{2}}}{2 \cdot 14}$

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

$x = \frac{- 9 \pm 5}{2 \cdot 14}$

Multiply $2$ by $14$ .

$x = \frac{- 9 \pm 5}{28}$

Step 4

The final answer is the combination of both solutions.

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