The two consecutive positive integers has a product of 272? What are the 4 integers?

Let a be the smaller of the two integers and let a+1 be the bigger of the two integers:
#(a)(a+1) = 272#, easiest way to solve this is to take the square root of 272 and round down:
#sqrt(272) = \pm16...#
16*17 = 272
Thus, the integers are -17,-16 and 16,17

If we multiply two consecutive numbers, #n and n+1#
we get #n^2+n#. That is we square a number and add one more on.

#16^2=256#

256+16=272

So our two numbers are 16 and 17

#color(blue)("A sort of cheat way")#

The two number are very close to each other so lets 'fudge' it

#sqrt(272) = 16.49...# so the first number is close to 16

Test #16xx17=272 color(red)(larr"First guess gets the prize!")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("The systematic way")#

Let the first value be #n# then the next value is #n+1#

The product is #n(n+1)=272#

#n^2+n-272=0#

Compare to: #ax^2+bx+c=0color(white)("ddd") -> color(white)("ddd") x=(-b+-sqrt(b^2-4ac))/(2a)#
In this case #x->n; color(white)("d")a=1; color(white)("d")b=1 and c=-272#

#n=(-1+-sqrt(1-4(1)(-272)))/(2(1))#

#n=-1/2+-sqrt(1089)/2#

#n=-1/2+-33/2# The negative is not logical so discard it

#n=-1/2+33/2 = 16#

The first number is 16 the second is 17