Quad-Rootika

Let us assume that $\sqrt{x-1}=t \implies x=t^2+1$

Now the given equation is equivalent to :

$\sqrt{t^2+t-4t}+\sqrt{t^2-9-6t}=1$

$\implies \sqrt{(t-2)^2}+\sqrt{(t-3)^2}=1$

$\implies |t-2|+|t-3|=1$

As we can see from the graph below, the above equation is satisfied for all values of y lying between $2$ and $3$ both inclusive.

So the solution is : $2 \leq t \leq 3 \implies 5 \leq x \leq 10$

So, total number of real values of $x$ lying between $5$ and $10$ both inclusive is infinite.

enjoy !