To find the minimum value of the expression $\frac{1}{2}b^2 + 5b - 3$, we can rewrite it in vertex form.
Completing the square, we have $\frac{1}{2}(b^2 + 10b) - 3 = \frac{1}{2}(b^2 + 10b + 25) - 3 - \frac{1}{2}(25)$.
Simplifying, we get $\frac{1}{2}(b + 5)^2 - \frac{31}{2}$.
The minimum value occurs when $(b + 5)^2$ is equal to 0, which happens when $b = \boxed{-5}$.
The answer is: -5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
For what real value of $b$ is the expression $\frac{1}{2}b^2 + 5b - 3$ minimized?

### Answer:
