positive real numbers

Question

 A positive real number is 6 less than another. If the sum of the squares of the two numbers is 38, then find the numbers.

Answer 1

The two numbers are y and y - 6

$y^2+(y^2-12y+36)=38$

$2y^2-12y-2=0$

$y^2-6y-1=0$

The quadratic formula gives an extraneous root and y = 3 + $\sqrt{10}$ = approximately 6.162 and

y - 6 = .162.