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

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

Add $9$ to both sides of the equation.

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

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 $b = 3$ .

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$