Find two consecutive positive even integers whose product is 288.

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,

$x(x+2)=288$

$\Rightarrow x^{2}+2 x-288=0$

$\Rightarrow x^{2}+18 x-16 x-288=0$

$\Rightarrow x(x+18)-16(x+18)=0$

$\Rightarrow(x+18)(x-16)=0$

$\Rightarrow x+18=0$ or $x-16=0$

$\Rightarrow x=-18$ or $x=16$

∴ x = 16           (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.