Geometric determination of a solution to the equation $x^{2}+px=q$

Let $x$ be a positive real number such that $x^{2}+px=q$.
We construct a figure as in the statement, such that the side length of the green square is $x$, and the width of the blue rectangles is $\frac{p}{2}$.

In doing so, the area of the green square plus the areas of the two blue rectangles is equal to $x^{2}+px$, which is equal to $q$ by assumption.

By adding the small red square of side length $\frac{p}{2}$, we obtain a large square of side length $x+\frac{p}{2}$. Adding all the areas, its area is
$$x^{2}+px+\left(\frac{p}{2}\right)^{2}=q+\frac{p^{2}}{4}.$$

Hence, taking square roots, the side length of the large square is:
$$x+\frac{p}{2}=\sqrt{q+\frac{p^{2}}{4}}.$$Finally, we obtain:
$$x=\sqrt{q+\frac{p^{2}}{4}}-\frac{p}{2}.$$

Note that this is indeed the strictly positive root that one would obtain using the discriminant.

Verification

Starting again from the equation $x^{2}+px-q=0$.
The discriminant is given by:
$$\Delta=p^2-4(-q)=p^2+4q>0.$$There are therefore two roots, given by:
$$x_1=\frac{-p-\sqrt{p^2+4q}}{2}=-\frac{p}{2}-\sqrt{\frac{p^2}{4}+q}, \quad x_2=-\frac{p}{2}+\sqrt{\frac{p^2}{4}+q}.$$This second root $x_2$ is exactly the $x$ found in the exercise; it is strictly positive since
$$\sqrt{\frac{p^2}{4}+q}>\sqrt{\frac{p^2}{4}}=\frac{p}{2}.$$