Find All Complex Solutions 9r^2+4r+4=0

$9 r^{2} + 4 r + 4 = 0$

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 $a = 9$ , $b = 4$ , and $c = 4$ into the quadratic formula and solve for $r$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$r = \frac{- 4 \pm \sqrt{16 - 4 \cdot 9 \cdot 4}}{2 \cdot 9}$

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

Multiply $- 4$ by $9$ .

$r = \frac{- 4 \pm \sqrt{16 - 36 \cdot 4}}{2 \cdot 9}$

Multiply $- 36$ by $4$ .

$r = \frac{- 4 \pm \sqrt{16 - 144}}{2 \cdot 9}$

Subtract $144$ from $16$ .

$r = \frac{- 4 \pm \sqrt{- 128}}{2 \cdot 9}$

Rewrite $- 128$ as $- 1 (128)$ .

$r = \frac{- 4 \pm \sqrt{- 1 \cdot 128}}{2 \cdot 9}$

Rewrite $\sqrt{- 1 (128)}$ as $\sqrt{- 1} \cdot \sqrt{128}$ .

$r = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{128}}{2 \cdot 9}$

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

$r = \frac{- 4 \pm i \cdot \sqrt{128}}{2 \cdot 9}$

Rewrite $128$ as $8^{2} \cdot 2$ .

Factor $64$ out of $128$ .

$r = \frac{- 4 \pm i \cdot \sqrt{64 (2)}}{2 \cdot 9}$

Rewrite $64$ as $8^{2}$ .

$r = \frac{- 4 \pm i \cdot \sqrt{8^{2} \cdot 2}}{2 \cdot 9}$

Pull terms out from under the radical.

$r = \frac{- 4 \pm i \cdot (8 \sqrt{2})}{2 \cdot 9}$

Move $8$ to the left of $i$ .

$r = \frac{- 4 \pm 8 i \sqrt{2}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$r = \frac{- 4 \pm 8 i \sqrt{2}}{18}$

Simplify $\frac{- 4 \pm 8 i \sqrt{2}}{18}$ .

$r = \frac{- 2 \pm 4 i \sqrt{2}}{9}$