Question 560148
Call the integers {{{n}}}, {{{n+1}}}, and {{{n+2}}}
given:
{{{ ( n+1 )*( n+2 ) = 56 }}}
{{{ n^2 + 3n + 2 = 56 }}}
{{{ n^2 + 3n - 54 = 0 }}}
Use quadratic formula
{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 1 }}}
{{{ b = 3 }}}
{{{ c = -54 }}}
{{{n = (-3 +- sqrt( 3^2-4*1*(-54) ))/(2*1) }}} 
{{{n = (-3 +- sqrt( 9 + 216 )) / 2 }}}
{{{n = (-3 +- sqrt( 225 )) / 2 }}} 
{{{n = ( -3 + 15 ) / 2 }}}
{{{n = 6 }}} ( I'll ignore the negative root of {{{225}}} )
{{{ n + 1 = 7 }}}
{{{ n + 2 = 8 }}}
The consecutive numbers are 6,7,and 8