solution: We complete the square.
  
  We have $(x+1.3)^2 = x^2 + (2.6)x + 1.69$, and so
  
  \begin{align*}
  x^2+(2.6)x+3.6 &= (x+1.3)^2 - 1.69 + 3.6 \\
  &= (x+1.3)^2 + 1.91.
  \end{align*}Therefore, $b=1.3$ and $c=1.91$, which gives us $b+c = \boxed{3.21}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  The quadratic $x^2+(2.6)x+3.6$ can be written in the form $(x+b)^2+c$, where $b$ and $c$ are constants. What is $b+c$ (as a decimal)?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 3.21
extra_info.question: The quadratic $x^2+(2.6)x+3.6$ can be written in the form $(x+b)^2+c$, where $b$ and $c$ are constants. What is $b+c$ (as a decimal)?
extra_info.split: test