Question 258156
Let x and x+1 be consecutive integers.
From above, we get
(i) {{{x(x+1) = 812}}}
distributing, we get
(ii) {{{x^2 + x = 812}}}
setting = 0, we get
(iii) {{{x^2 + x - 812 = 0}}}
factoring, we get
(iv) {{{(x+29)(x-28) = 0 }}}
solving, we get
x = -29
or
x = 28.
So our two options are:
(-29, -28)
(28, 29)