Algebra

The first step is to combine like terms in the original expression in the question to make the quadratic clearer:

$x^2 + 6x + 14$

Now we must open the parenthesis for (x + u)^2.

$(x + u)^2$ = $x^2 + 2xu + u^2$

$u$ is a constant so $u^2$ is also a constant, meaning that $2xu$ is the only term that can equal $6x$ in the expression in the question. 

Therefore the constant $u = 3$. 

If $u$ is 3, then $u^2$ would be 9, making the new expression: 

$ax^2 + 6ax + 9a + v$

(don't forget about the variable $a$ that multiplies the outcome of opening the parenthesis for (x + u)^2)

If you take a look at the expression the problem gave to us, the coefficient of $x^2$ was 1, so that means the constant $a$ is also 1.

The new expression would look like this and we can get the last constant we need:

$x^2 + 6x + 9 + v$

Now we see there is a difference of 5 from the expression above and the expression in the problem, so $v = 5$.

Therefore the answer for this questions is:

$1(x + 3)^2 + 5$

where $a = 1$, $u = 3$, and $v = 5$