We have to prove that a^2 - 4a + b^2 + 10b + 29>=0, for real values of a and b.

a^2 - 4a + b^2 + 10b + 29

=> a^2 - 4a + 4 + b^2 + 10b + 25

=> (a - 2)^2 + (b + 5)^2

The sum of squares of real numbers is always positive.

**This proves that a^2 - 4a + b^2 + 10b + 29 >= 0**