yikes, no. You are told those are the roots of some unspecified quadratic equation.
In order for them to be real we need the argument to that square root to be non-negative.
Otherwise that bit would contribute an imaginary part.
Ok so that means
\(2p-1\geq 0\\ 2p\geq 1\\ p \geq \dfrac 1 2\)
so there's not a single value of p that causes the roots to be real but a half closed interval of them.
For two equal valued roots we simply have the bit in the square root be equal to 0.
\(2p-1=0\\ p = \dfrac 1 2\)
For rational but unequal roots the bit in the square root has to be a perfect square.
\(2p-1 = k^2,~k \in \mathbb{N}\\ p = \dfrac{k^2+1}{2},~k \in \mathbb{N}\)
for example p = 5 provides 2 rational but unequal roots.
