Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!

Set $a=\sqrt{x-4}$ and $b=\sqrt{x-5}$ for $-2\leq x\leq +2$. Equivalently we have $a^2=x-4$,  $b^2={x-5}$ for $-2\leq x\leq +2$. It's implies for $-2\leq x\leq +2$,
\begin{align}
a-b=-1 \implies & (a-b)(a+b)=-1\cdot(a+b)\\
       \implies & a^2-b^2=-(a+b)\\
       \implies & (x-4)-(x-5)=-(a+b)\\
       \implies & -1=-(a+b)\\
       \implies & a+b=+1.\\
\end{align}
The same calculations and similar implications $a+b=+1\implies a-b=+1$. Then $ x $ is a root of $a+b=-1$, if only if, 
\begin{cases}
a+b=&+1\\
a-b=&-1\\
\end{cases}
Here, $a=0$ and $b=1$ implies $a^2=x-4=0$ and $b^2={x-5}=1$. In both cases, $x\in\!\!\!\!\!/[-2,+2]$.
So the equation 
$$
\sqrt{x-4}-\sqrt{x-5}+1=0 
$$
has no solution.
