Question 1118597
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<pre>
x^2 + (x+6)^2 = 38


x^2 + x^2 + 12x + 36 = 38


2x^2 + 12x = 2 


x^2 + 6x = 1


(x+3)^2 = 10


x = - 3 +/- {{{sqrt(10)}}}.


Since we want a positive number, the answer is


x = -3 + sqrt(10).


The two real numbers are  {{{-3 + sqrt(10)}}}  and  {{{3 + sqrt(10)}}}.
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Solved.