Question 405164
let two integers be {{{x}}} and {{{y}}}

if we have two consecutive integers, then {{{x}}} and {{{y=x+1}}}


their sum is 19 less than their product, so

{{{x + (x+1) + 19 = x(x+1)}}}.....solve for {{{x}}}

{{{2x+20 = x^2+x}}}

{{{0 = x^2+x -2x -20}}}

{{{0 = x^2 -x -20}}}

{{{ x^2 -x -20=0}}}.....find roots if {{{a=1}}},{{{b= -1}}} and {{{c= -20}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-(-1) +- sqrt( (-1)^2-4*1*(-20) ))/(2*1) }}}


{{{x = (1 +- sqrt( 1+80 ))/2 }}}


{{{x = (1 +- sqrt( 81 ))/2 }}}


{{{x = (1 +- 9)/2 }}}


{{{x1 = (1 + 9)/2 =10/2=5}}}

{{{x2 = (1 - 9)/2 =-8/2=-4}}}

now find 

{{{(x1+1)=5+1=6}}}

{{{(x2+1)=-4+1=-3}}}

so, you have two pairs of consecutive numbers: 

first pair: {{{x=5}}},and {{{(x+1)=6}}}

second  pair:{{{x=-4}}}, and {{{(x+1)=-3}}}





check:

{{{x + (x+1) + 19 = x(x+1)}}}

{{{5+ (5+1) + 19 = 5(5+1)}}}

{{{5+ 6 + 19 = 30}}}

{{{30 = 30}}}


or, other pair

{{{x + (x+1) + 19 = x(x+1)}}}

{{{-4+ (-3) + 19 = -4(-3)}}}

{{{-7 + 19 = 12}}}

{{{12 = 12}}}