Question

one number is 6 more than anouther number. the sum of their squares is 90

Answer 1

Let x = greater number
      y = smaller  number

(1) $x = y + 6$
(1) $y = x-6$

(2) $x^{2} + y^{2} = 90$
We'll substitute y in (1) to (2)
(2) $x^{2} + (x-6)^{2} = 90$
     $x^{2} + (x^{2} - 12x + 36) = 90$
     $x^2 + x^2 -12x + 36 - 90 = 0$
     $2x^2 - 12x -54 = 0$
     $2(x^2 - 6x - 27) = 0$
     $2(x - 9)(x + 3) = 0$
 x - 9 = 0        or      x + 3 = 0
      x = 9                        x = -3
and
 y = x - 6                  y = x - 6
 y = 9 - 6                  y = -3 - 6
 y = 3                        y = -9

Therefore, the two numbers can be 9 and 3 or -3 and -9.