Solve Using the Quadratic Formula 7x^2-2x-9=0

$7 x^{2} - 2 x - 9 = 0$

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 $a = 7$ , $b = - 2$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (7 \cdot - 9)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 7 \cdot - 9}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 9$ .

Multiply $- 4$ by $7$ .

$x = \frac{2 \pm \sqrt{4 - 28 \cdot - 9}}{2 \cdot 7}$

Multiply $- 28$ by $- 9$ .

$x = \frac{2 \pm \sqrt{4 + 252}}{2 \cdot 7}$

Add $4$ and $252$ .

$x = \frac{2 \pm \sqrt{256}}{2 \cdot 7}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{2 \pm \sqrt{16^{2}}}{2 \cdot 7}$

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

$x = \frac{2 \pm 16}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{2 \pm 16}{14}$

Simplify $\frac{2 \pm 16}{14}$ .

$x = \frac{1 \pm 8}{7}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{9}{7}, - 1$