Solving the quadric equation: $\frac{-U\pm \sqrt{U^2-4V}}{2}$ where $U,V$~$\mathbf{U}(-1,1)$ independently. I'm given two uniform random variables $V,U \sim\mathbf{U}(-1,1)$.
I also get the function: $$h(s)=s^2+Us+V$$
I'm interested to answer queations like for which values $h(s)$ has only: zero /one/ two roots and what is the probability for that.
So of course I have to use this: $$s_{1,2}=\frac{-U\pm \sqrt{U^2-4V}}{2}$$
One root: $U^2=4V$, now correct me if I'm wrong, but since there is no inequality sign this is not an accumulating event and therefore its probability is $0$? Now I have the feeling I'm terribly wrong but I have no way to prove myself wrong.
Two roots: $$U^2>4V \rightarrow P(U^2>4V)=1-P(U^2<4V) = 1-P(-2\sqrt{V}<U<2\sqrt{V})$$ And again I get to a phrase I'm not 100% sure is correct, let along I'm not sure how to calculate. Can anyone hint / guide how to proceed from here? what kind of method should I generally use in problems like this?
 A: 
since there is no inequality sign this is not an accumulating event and therefore its probability is $ 0$?

The conclusion is right. The intuition is also basically right, though the expression "accumulating event" is unknown to me. The correct term would be "a zero measure" event.
Basically: because $U,V$ are uniform and independent on the unit square, the probability of any event $E$ that depends on some (in)equality on $U,V$, is given by the area corresponding to that event on the $U,V$ plane. Because in this case the condition is $V=U^2/4$ , the region is a curve (a parabola) , and its area is zero.
In general (roughly) : inequalities like $a < g(U,V) < b$ will give regions corresponding to areas (hence positive probability... unless the region doesn't intersect the unit square) and equalities like $g(U,V)=0$ will give you a curve (hence zero probability... unless the equation is degenerate , like $0 U + 0 V = 0$).
The second part corresponds to the inequality case: to compute $P(U^2<4V)$ you need to integrate the joint density over that region, which in this case correponds to computing its area:
$$P(U^2<4V)=\int_{u^2<4v}f_{U,V}(u,v) du dv $$
To compute the integral you need to graph the region and compute its area by using the proper integrals.
