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

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

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

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

Subtract $9$ from both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $6 x - x^{2} - 9$ .

Reorder $6 x$ and $- x^{2}$ .

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

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $6 x$ .

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

Rewrite $- 9$ as $- 1 (9)$ .

$- (x^{2}) - (- 6 x) - 1 \cdot 9 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 6 x)$ .

$- (x^{2} - 6 x) - 1 \cdot 9 = 0$

Factor $- 1$ out of $- (x^{2} - 6 x) - 1 (9)$ .

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

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Divide each term in $- {(x - 3)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 3)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 3)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 3)}^{2}}{1} = \frac{0}{- 1}$

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

${(x - 3)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

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

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$