How to find how many real roots of an equation?

Given that the roots of $ax^2+bx+c$ are both positive, and real, let the roots be $x=\alpha, \beta>0$ which follows from the condition. Note that the quadratic with the absolute value can be changed into $$ax^2+b|x|+c=0 \iff a|x|^2+b|x|+c=0$$ So we have$$|x|=\alpha, \beta$$ is a root. So there are four roots, $\alpha$, $-\alpha$, $\beta$, $-\beta$.

The solutions to this equation are the solutions to any of the equations : $ax^2+bx+c=0$ and $ax^2-bx+c=0$.

Hence the solutions to the following can be (at most $4$ values) : $$\frac { \pm b \pm \sqrt {b^2-4ac}}{2a} $$

Hope it helps.