If both $x+10$ and $x-79$ are perfect squares, what is $x$? 
If both $x+10$ and $x-79$ are perfect squares, what is $x$?

That is: 
$x+10=$perfect square  (or can be square such that the result of it squared is a positive integer)
$x-79=$perfect square  (or can be square such that the result of it squared is a positive integer)
I have no idea how to solve this.I tried inequality like
$$x+10>x-79$$
$$x-x>-79-10$$
$$0>-89$$
Note: You are not allowed to use trial and error or guess and check to solve the question.

Assuming you are looking for positive integers...
Consecutive squares differ by consecutive odd amounts. The first differences to $\{0,1,4,9,16,\ldots\}$ are $\{1,3,5,7,\ldots\}$.
You have two squares that are $89$ apart. They may or may not be consecutive. But for some string of consecutive odd numbers, the sum must be $89$.
$$
\begin{align}
(2k+1)+\left(2(k+1)+1\right)\cdots+\left(2(k+h)+1\right)&=89\\
2k(h+1)+h(h+1)+h+1&=89\\
(2k+h+1)(h+1)&=89\\
\end{align}$$
Since $89$ is prime, either $h=0$ and $k=44$, or well, there is no other possibility.
So $k=44$ and there is only one consecutive odd number in our sequence, meaning the two squares must be consecutive squares. We must be dealing with the difference between $44^2$ and $45^2$. So $x=45^2-10$.
