Q:

Completing the Square (Solving for $x$ with fractions)

Im working through some completing the square questions but this question I'm currently working on has a few fractions which seem to be tripping me up. Some advice on where I have gone wrong would be greatly appreciated.
Q) Complete the Square to solve for $x$:  $\qquad x^2 - 5x - 6 = 0$
So far I have:
\begin{align}
x^2 - 5x &= 6  \\
x^2 - 5x + \frac{25}4  &= \frac{49}4   \\
\left(x - \frac52\right)\left(x - \frac52\right) &= \frac{49}4  \\ 
\left(x - \frac52\right)^2 = \frac{49}4    
\end{align}
Take the square root:
\begin{align}
x - \frac52 &= \pm \frac72   \\
x_1 &=  \frac72 + \frac52 =  \frac{12}2 = 6 \\
x_2 &= -\frac72 + \frac52 = -\frac22    = -1
\end{align}

A:

What you have done is correct,
$$ 0= x^2-5x-6 = \left( x-\frac{5}{2} \right)^2 - \frac{5^2}{2^2}-6 \\
= \left( x-\frac{5}{2} \right)^2 - \frac{49}{4}, $$
so
$$ \left( x-\frac{5}{2} \right) = \pm \frac{7}{2},  $$
so $x=-1$ or $x=6$. You can check these in the original equation:
$$ (-1)^2-5(-1)-6 = 0\\
6^2-5 \times 6 -6 = 0. $$
Also, it is possible to spot the factorisation as
$$ (x+1)(x-6). $$

