Find the real numbers a,b,c if a^2+b^2+c^2-4a-6b+10c+38=0?

We'll re-group the terms to emphasize the fact that we'll
have to complete a number of squares:

(a^2 - 4a) + (b^2 -
6b) + (c^2 + 10c) + 38 = 0

We'll have to complete the
squares inside the brackets:

(a^2 - 4a + 4) + (b^2 - 6b +
9) + (c^2 + 10c + 25) - 4 - 9 - 25 + 38 = 0

(a^2 - 4a + 4)
+ (b^2 - 6b + 9) + (c^2 + 10c + 25) - 38 + 38 = 0

We'll
eliminate like terms and we'll recognize the perfect
squares:

(a-2)^2 + (b-3)^2 + (c+5)^2 =
0

The sum of the squares cannot be zero, unless the value
of each square is zero.

a - 2 = 0 <=> a =
2

b - 3 = 0 <=> b =
3

c + 5 = 0 <=> c =
-5

The requested values of the numbers a,b,c,
for the given identity to hold, are: a = 2 , b = 3 , c =
-5.