Conditional Minimum

Since, the given quadratic does not have 2 distinct, real roots, it's roots must be repeated or imaginary.

Thus, the discriminant must fulfill the following,

$b^2-20a \le 0$ $\therefore b^2 \le 20a$

Subtracting $10b$ from both sides,

$b^2-10b \le 20a-10b$ $\therefore b(b-10) \le 10(2a-b)$ $\therefore \frac {b(b-10)}{10} \le 2a-b$

Therefore, if we minimise $b(b-10)$ we minimise $2a-b$

We know that $b(b-10)$ has it's minimum value of $(-25)$ at $b=5$ (i.e at the vertex of it's graph).

$\therefore \frac {-25}{10} \le 2a-b$ $\therefore \frac {-5}{2} \le 2a-b$

Therefore, $(2a-b)_{min}=-2.5$