Two number differ by 3 and their product is 504.

Solution:

Let two required numbers be $x$ and $(x+3)$

Then according to question

$x(x+3)=504$

$x^{2}+3 x-504=0$

$x^{2}+24 x-21 x-504=0$

$x(x+24)-21(x+24)=0$

$(x+24)(x-21)=0$

$(x+24)=0$

$x=-24$

Or

$(x-21)=0$

$x=21$

Since, x being a number,

Therefore,

When $x=-24$ then

$x+3=-24+3$

$=-21$

And when $x=21$ then

$x+3=21+3$

$=24$

Thus, two consecutive number be either 21,24 or $-21,-24$