Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me? 
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help

$\sqrt{2x-5} - \sqrt{x-1} = 1$
Let $\sqrt{2x-5} + \sqrt{x-1} = y$
Multiplying, we get
$(2x-5) - (x-1) = y$
$y = x - 4$
\begin{align}
   \sqrt{2x-5} + \sqrt{x-1} &= x - 4 \\
   \sqrt{2x-5} - \sqrt{x-1} &= 1 & \text{subtract}\\
\hline
   2\sqrt{x-1} &= x-5 \\
   4x-4 &= x^2 - 10x + 25 \\
   x^2 -14x + 29 &= 0 \\
   x &= 7 \pm 2 \sqrt 5
\end{align}
