Answer
$x = 2 \quad \ \textbf{or}\ \ \ x= 7 $
Work Step by Step
First transfer the constant to the right of the equal sign proceed with completing the square in line 1 of the equation by dividing the coefficient of the $9x$ term by $2$ and squaring it and adding the result on both sides. $$
\begin{aligned}
x^2-9x +14& = 0 \\
x^2-9x & = -14 \\
x^2-9x +\left(\frac{9}{2}\right)^2 & =-14+\left(\frac{9}{2}\right)^2 \\
x^2-9x +4.5^2 & =-14+4.5^2 \\
(x-4.5)^2 & = 6.25
\end{aligned}
$$ Take the square root: $$
\begin{array}{rl}
(x-4.5)^2 & = 6.25 \\
x-4.5 & = \pm \sqrt{6.25} \\
x-4.5 & = \pm 2.5 \\\\
x-4.5=2.5 &\ \textbf{or}\ \ \ x-4.5=-2.5\\
x=4.5+2.5 & \ \textbf{or}\ \ \ x= 4.5-2.5 \\
x = 7 & \ \textbf{or}\ \ \ x = 2
\end{array}
$$ Check $$\begin{aligned}
7^2-9(7) +14& \stackrel{?}{=} 0 \\
0& = 0\checkmark
\end{aligned}$$ $$\begin{aligned}
2^2-9(2) +14& \stackrel{?}{=} 0 \\
0& = 0\checkmark.
\end{aligned}$$ The solution is:
$$x = 7 \quad \ \textbf{or}\ \ \ x= 2 $$
