The product of two successive integral multiples of 5 is 300.

Solution:

Let the successive integer multiples of 5 be $5 x$, and $5(x+1)$

Then according to question

$5 x \times 5(x+1)=300$

$25\left(x^{2}+x\right)=300$

$x^{2}+x=12$

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

$x^{2}-3 x+4 x-12=0$

$x(x-3)+4(x-3)=0$

$(x-3)(x+4)=0$

Therefore,

$(x-3)=0$

$x=3$

Or

$(x+4)=0$

$x=-4$

When $x=3$ then integer

$5 x=5 \times 3$

$=15$

$5(x+1)=5(3+1)$

$=5 \times 4$

 

$=20$

And when $x=-4$ then integer

$5 x=5 \times-4$

$=-20$

$5(x+1)=5(-4+1)$

$=5 \times(-3)$

$=-15$

Thus, three consecutive positive integer be 15,20 or $-20,-15$