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

$- 7 d^{2} + 2 d + 9 = 0$

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 $d$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 7$ .

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

Multiply $28$ by $9$ .

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

Add $4$ and $252$ .

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

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

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

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

$d = \frac{- 2 \pm 16}{2 \cdot - 7}$

Multiply $2$ by $- 7$ .

$d = \frac{- 2 \pm 16}{- 14}$

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

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

Step 4

The final answer is the combination of both solutions.

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