We can solve this problem by using #x# for one integer and, since they are consecutive numbers, #x+1# for the other.

We know that #x * (x+1) = 72#.

This is equivalent to #x^2 + x = 72#. After rearranging the equation, we get

#x^2 + x - 72 =0# This is now a tipical quadratic equation #ax^2 + bx + c =0#, for which

#Delta# = #(b^2 -4ac)# #-># #Delta = 1^2 - 4 * 1 *(-72) = 289#

Now, we know that #x_(1,2) = (-b +- sqrt(Delta))/(2a) = (-1 +- sqrt(289))/2#. Therefore, we have

#x_1 = (-1 -17)/2 = -9# and #x_2 = (-1 +17)/2 = 8#

So, there are four numbers that satisfy the initiaul equation: for #x=-9# we have #-9# and #(-9 + 1) = -8#, and for #x = 8# we have #8# and #(8 + 1) = 9#.

Therefore, #-9 * (-8) = 72# and #8 * 9 = 72#.