Key idea. Complete one square in each variable.

1. For the x terms, \(x^2-8x=(x-4)^2-16\). For the y terms, \(y^2+2y=(y+1)^2-1\).

2. Substitute these expressions and collect constants:

   \[x^2-8x+y^2+2y+20=(x-4)^2+(y+1)^2+3.\]

3. The squares are non-negative, so the expression is at least 3.

4. We can make both squares zero simultaneously by choosing \(x=4\), \(y=-1\). The value 3 is therefore actually reached.

Remember. Finding a lower bound is only half an optimisation proof: also show that it can be attained.