The sum of two natural numbers is 28 and their product is 192. Find the numbers.

Let the required numbers be x and (28 − x).

According to the given condition,

$x(28-x)=192$

$\Rightarrow 28 x-x^{2}=192$

$\Rightarrow x^{2}-28 x+192=0$

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

$\Rightarrow x(x-16)-12(x-16)=0$

$\Rightarrow(x-12)(x-16)=0$

$\Rightarrow x-12=0$ or $x-16=0$

$\Rightarrow x=12$ or $x=16$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.