The equation $x^2 + p x + q = 0$ has real roots, therefore $p^2 - 4q \ge 0$. Let one of the roots be $\frac{-p \pm \sqrt{p^2 - 4q}}{2}$.
Then
$$\begin{aligned} \frac{-p \pm \sqrt{p^2 - 4q}}{2} &\ge \frac{4q - (p + a)^2}{4a} \\ &\Leftrightarrow 2a(-p \pm \sqrt{p^2 - 4q}) \ge 4q - (p + a)^2 \\ &\Leftrightarrow -2a p \pm 2a \sqrt{p^2 - 4q} \ge 4q - p^2 - 2a p - a^2 \\ &\Leftrightarrow p^2 - 4q \pm 2a \sqrt{p^2 - 4q} + a^2 \ge 0 \\ &\Leftrightarrow (\sqrt{p^2 - 4q} \pm a)^2 \ge 0. \end{aligned}$$