Two number differ by 4 and their product is 192.

Solution:

Let two required numbers be $x$ and $(x+4)$

Then according to question

$x(x+4)=192$

$x^{2}+4 x-192=0$

$x^{2}+16 x-12 x-192=0$

$x(x+16)-12(x+16)=0$

$(x+16)(x-12)=0$

$(x+16)=0$

$x=-16$

Or

$(x-12)=0$

$x=12$

Since, x being a number,

Therefore,

When $x=-16$ then

$x+4=-16+4$

$=-12$

And when $x=12$ then

$x+4=12+4$

$=16$

Thus, two consecutive number be either 12,16 or $-16,-12$