Question 5153
Let first number be x. Since both add to 10, the other number must be 10-x.


other info.... {{{x^2 + (10-x)^2 = 58}}}

{{{x^2 + 100 - 20x + x^2 = 58}}}

{{{2x^2 - 20x + 100 = 58}}}

{{{2x^2 - 20x + 42 = 0}}}
{{{x^2 - 10x + 21 = 0}}}
(x-7)(x-3) = 0

so x-7=0 or x-3=0


so x is 3 or 7...the 2 are interchangeable results ie the 2 numbers are 3 and 7.


jon.