Question 701482
Let {{{ N }}} be one of the numbers



"One number exceeds another by 5":
then the other is {{{ N+5 }}}



"The sum of their squares is 157":
{{{ N^2 + (N+5)^2 = 157 }}}
{{{ N^2 + N^2 + 10N + 25 = 157 }}}
{{{ 2N^2 + 10N -132 = 0 }}}
{{{ N^2 + 5N -66 = 0 }}}
{{{ (N-6)(N+11) = 0 }}}
This quadratic has two solutions: {{{ N=6 }}} and {{{ N=-11 }}}


Consider N=6:
1st number=N=6, 2nd number=N+5=11


Consider N=-11:
1st number=N=-11, 2nd number=N+5=-6