Question 954690

Let's call one of the numbers {{{x}}}. Since the numbers are {{{consecutive}}}, the next number has to be {{{one}}} more, so we'll call it {{{x+1}}}. 

The product of these numbers is {{{x(x+1)}}}.

given: the product of two consecutive numbers is {{{930}}}

so,  we have {{{x(x+1)=930}}}..........solve for {{{x}}}

{{{x(x+1)=930}}}

{{{x^2+x=930}}}

{{{x^2+x-930=0}}}

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

{{{x = (-1 +- sqrt(1^2-4*1*(-930) ))/(2*1) }}} 

{{{x = (-1 +- sqrt(1+3720 ))/2 }}} 

{{{x = (-1 +- sqrt(3721 ))/2 }}} 

{{{x = (-1 +- 61)/2 }}}

solutions:

{{{x = (-1 + 61)/2 }}}

{{{x =  60/2 }}}

{{{highlight(x = 30) }}}-> first number

{{{x = (-1 -61)/2 }}}

{{{x =  -62/2 }}}

{{{highlight(x = -31) }}}-> second number

then first consecutive number is {{{highlight(x+1 = 31) }}} and

second consecutive number is {{{highlight(x+1 = -30) }}}

so, you have two pairs of consecutive numbers that have a product {{{930}}} and they are:
1.{{{highlight( 30) }}} and {{{highlight(31) }}}

2.{{{highlight(-31) }}} and {{{highlight(-30) }}}