Find all real number satisfying $10^x+11^x+12^x = 13^x+14^x$ 
Find all real number $x$ satisfying $10^x+11^x+12^x = 13^x+14^x$    

My Work
Dividing by $13^x$ we get 
$$\left( \frac{10}{13} \right)^x + 
\left( \frac{11}{13} \right)^x + 
\left( \frac{12}{13} \right)^x
= 1 + \left( \frac{14}{13} \right)^x$$
The LHS is a decreasing function of $x$ and the RHS is an increasing function of $x$. So there is only one intersection in their graph. I am looking for a formal way to find the root. I know that $x=2$ works. But how to formally find this root?

I just solved my problem. A well known fact that 
$$a^2 +(a+1)^2 + \cdots + (a+k)^2 = (a+k+1)^2 +(a+k+2)^2+ \cdots (a+2k)^2$$
Satisfies when $a = k(2k+1)$.
So in my question $a = 2(2\cdot2+1)$, So without any thinking $x=2$ is an answer.
This helps suppose the equation was - 
$$55^x+56^x+57^x+58^x+59^x+60^x=61^x+62^x+63^x+64^x+65^x$$
Then it is hard to just substitute $x = 1,2,3,..$ and check? But using this fact without any thinking we can say $x=2$, as $55 = 5(5\cdot2+1)$. 
