Solution of equation If $f(x) = x^2 - 2ax + a(a+1)$ , $f:[ a, \infty] \to [a,\infty]$ . If one of the solution of the equation $f(x)=f^{-1}(x)$ is $5049$ , then what may be the other solution ?
My WORK:
I found the inverse, but when I equate them , it is very difficult to solve . Please help me with some method to solve it.

Your method of equating the function and its inverse directly is perfectly alright.  
$f(x) = x^2 - 2ax + a^2 + a$
or  $f(x) = (x-a)^2 + a$
or  $f(x) - a = (x-a)^2$
$x = a \pm \sqrt{f(x) - a}$  
Here the negative sign is rejected as $x \in [a,\infty]$  
So, we have $f^{-1}(x) = a + \sqrt{x - a}$  
Now, $f(x) = f^{-1}(x)$
$(x-a)^2 + a = a + \sqrt{x-a}$
$(x-a)^2 = \sqrt{x-a}$  
$(x - a) ^ {\frac{1}{2}} ((x - a) ^ {\frac{3}{2}} - 1) = 0$  
Therefore,
$x - a = 0$ or $(x - a)^{\frac{3}{2}} = 1$
$x = a$, or  $x = a + 1$
