Find the Roots (Zeros) 9x^2-6x=-9

$9 x^{2} - 6 x = - 9$

Step 1

Add $9$ to both sides of the equation.

$9 x^{2} - 6 x + 9 = 0$

Step 2

Factor $3$ out of $9 x^{2} - 6 x + 9$ .

Factor $3$ out of $9 x^{2}$ .

$3 (3 x^{2}) - 6 x + 9 = 0$

Factor $3$ out of $- 6 x$ .

$3 (3 x^{2}) + 3 (- 2 x) + 9 = 0$

Factor $3$ out of $9$ .

$3 (3 x^{2}) + 3 (- 2 x) + 3 (3) = 0$

Factor $3$ out of $3 (3 x^{2}) + 3 (- 2 x)$ .

$3 (3 x^{2} - 2 x) + 3 (3) = 0$

Factor $3$ out of $3 (3 x^{2} - 2 x) + 3 (3)$ .

$3 (3 x^{2} - 2 x + 3) = 0$

Step 3

Divide each term in $3 (3 x^{2} - 2 x + 3) = 0$ by $3$ and simplify.

Divide each term in $3 (3 x^{2} - 2 x + 3) = 0$ by $3$ .

$\frac{3 (3 x^{2} - 2 x + 3)}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (3 x^{2} - 2 x + 3)}{3} = \frac{0}{3}$

Divide $3 x^{2} - 2 x + 3$ by $1$ .

$3 x^{2} - 2 x + 3 = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

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

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

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

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

Step 6

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 3 \cdot 3$ .

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12 \cdot 3}}{2 \cdot 3}$

Multiply $- 12$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 36}}{2 \cdot 3}$

Subtract $36$ from $4$ .

$x = \frac{2 \pm \sqrt{- 32}}{2 \cdot 3}$

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 3}$

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

$x = \frac{2 \pm i \cdot \sqrt{32}}{2 \cdot 3}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{2 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 3}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (4 \sqrt{2})}{2 \cdot 3}$

Move $4$ to the left of $i$ .

$x = \frac{2 \pm 4 i \sqrt{2}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{2 \pm 4 i \sqrt{2}}{6}$

Simplify $\frac{2 \pm 4 i \sqrt{2}}{6}$ .

$x = \frac{1 \pm 2 i \sqrt{2}}{3}$

Step 7

Find the Roots (Zeros) 9x^2-6x=-9

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