In which range a must lie?

Using Calculus...

$f(x) = x^2 + ax + b$ , so $f'(x) = 2x + a.$

We're told $f(x)$ has a minimum at $x = 2a + 5b$ , thus $f'(2a + 5b) = 0$ .

Now, $f'(2a + 5b) = 4a + 10b + a = 5a + 10b$ . So, $f'(2a + 5b) = 0$ implies $5a + 10b = 0$ , i.e. $a = -2b$ .

We're told that the graph of f intersects the $x$ -axis. Since the leading coefficient is positive, this implies that our minimum must be zero or negative.
The minimum is at $x_0 = 2a + 5b$ . But we can simplifying this buy substituting $a = -2b$ , and thus $x_0 = -4b + 5b = b$ . Going back to the original function, we see $f(b) = b^2 + ab + b = b^2 - 2b^2 + b = b - b^2$ .
Thus $b - b^2 \leq 0$ , i.e. $b^2 \geq b$ . This inequality holds for $b \geq 1$ and for $b \leq 0$ . Substituting again, our solution space is $-a/2 \geq 1$ or $-a/2 \leq 0$ , i.e. $a \leq -2$ or $a \geq 0$ . In interval notation, this means $a \in (-\infty,-2] \cup [0,\infty)$