Question 409556
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Hi
one of the integers is five less than seven times the other integer.
Let x and (7x-5)represent the intergers
Question states***
 x(7x-5) = 408
   7x^2 - 5x -408 = 0
Solving for x

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (5 +- sqrt( 11449))/(14) }}}
{{{x = (5 +- 107)/14}}}
   x = 112/14 = 8  and x = -102/14 (extraneous solution as is a non-integer)  
Integers are 8 and 51 (7*8-5)

CHECKING our Answer***
   8*51 =408