Question 712953
Consecutive integers are integers that follow in sequence, each number being {{{1}}} more than the previous number, represented by {{{n}}}, {{{n +1}}}, {{{n + 2}}},{{{ n + 3}}}, ..., where {{{n}}} is any integer. 

in your case we have two {{{negative}}} consecutive integers; so, let them be  {{{n}}} and {{{n +1}}}

their the product is {{{56}}}; so, we have

{{{n(n +1)=56}}}

{{{n^2 +n=56}}}

{{{n^2 +n-56=0}}}.......use quadratic formula to solve for {{{n}}}

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

{{{n = (-1 +- sqrt( 1^2-4*1*(-56) ))/(2*1) }}} 

{{{n = (-1 +- sqrt( 1+224))/2 }}} 

{{{n = (-1 +- sqrt( 225))/2 }}} 

{{{n = (-1 +- 15)/2 }}}...we will need only {{{negative}}} solution for {{{n}}}

 {{{n = (-1 - 15)/2 }}}

 {{{n = (-16)/2 }}}

 {{{highlight(n = -8 )}}}...first number

{{{n +1=-8+1=highlight(-7)}}}...second number