Find All Complex Solutions 6x=9+x^2

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

Move all the expressions to the left side of the equation.

Subtract $9$ from both sides of the equation.

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

Subtract $x^{2}$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 1$ , $b = 6$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (- 1 \cdot - 9)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 1 \cdot - 9}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 6 \pm \sqrt{36 + 4 \cdot - 9}}{2 \cdot - 1}$

Multiply $4$ by $- 9$ .

$x = \frac{- 6 \pm \sqrt{36 - 36}}{2 \cdot - 1}$

Subtract $36$ from $36$ .

$x = \frac{- 6 \pm \sqrt{0}}{2 \cdot - 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{- 6 \pm \sqrt{0^{2}}}{2 \cdot - 1}$

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

$x = \frac{- 6 \pm 0}{2 \cdot - 1}$

$- 6$ plus or minus $0$ is $- 6$ .

$x = \frac{- 6}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 6}{- 2}$

Divide $- 6$ by $- 2$ .

$x = 3$

The final answer is the combination of both solutions.

$x = 3$ Double roots