## 1 Answer

The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]for a quadratic equation in the form \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -6\), and \(c = 9\).

So\[x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)} = \frac{6 \pm \sqrt{36 - 36}}{2} = \frac{6 \pm 0}{2} = 3.\]Therefore, the roots of the equation are both \(x = 3\).