$x$

Given in the question $\rightarrow$ $100+x$ is a perfect square and $168+x$ is a perfect square.

Let us subtract: $(168+x)-(100+x)=68$ .

So, the difference of two squares is $68$ .

$\Rightarrow\space m^2-n^2=68$ for some positive $m$ and $n$ .

$\Rightarrow\space (m+n)(m-n)=68\times1\space or\space 34\times2\space or\space 17\times4$

Solving, we get: (fractional values for $m$ and $n$ are discarded)

$m=18$

$n=34-18=16$

$\therefore\space 168+x=m^2=18^2=324\Rightarrow\space x=324-168=156$ or

$100+x=n^2=16^2=256\Rightarrow\space x=256-100=156.$