Probablgebra!!

The first equation could be rewrite as $(x+\frac{p}{2})^2 = \frac{p^2}{4}-\frac{p}{4}-\frac{1}{4}$ .

In order for this to have real roots, the RHS must be non negative. Apply that and rewrite we have

$(\frac{p}{2}-\frac{1}{4})^2 \geq \frac{9}{16}$

or $\frac{p}{2}-\frac{1}{4}\geq\frac{3}{4}\text{ or }p\geq2$

Because we choose p random from [0,5] so we have probability of $p\geq2\text{ is } \frac{5-2}{5}\text{ so our answer is }\boxed{8}$

Note that there is a small case where $\frac{1}{4}-\frac{p}{2}\geq\frac{3}{4}$ , but that could not be satisfy with p in [0,5]

I also did the same b^2-4ac>=0... p>=2 or p <= -1. Thus 2-5 is the answer.
Probability (5-2)/5=3/5... answer=3+5=8