The sum of the two numbers is $29$ and the difference of their squares is $145.$ The difference between the numbers is:

The correct option is C: $5$

Let the two numbers be $x$ and $y,$ where $x>y$.

It is given that the sum of two numbers is 29.

$x+y=29\dots \dots \left(i\right)$

And, the difference of their square is 145.

$x^2-y^2=145$

$\Rightarrow (x+y)(x-y)=145$ [$\because (a^2-b^2)=(a+b)(a-b)$]

$\Rightarrow (x-y)\times 29=145$ [Using eq.(i)]

$\Rightarrow (x-y)=\frac{145}{29}$

$\therefore (x-y)=5$

Hence, the difference between the two numbers is %5.$