Step 1: Set up the discriminant condition.
For $x^2+(2m+1)x+m=0$ to have equal roots, the discriminant must be zero. Here $a=1$, $b=2m+1$, $c=m$. So we need $D = b^2 - 4ac = 0$.
Step 2: Compute the discriminant.
\[ D = (2m+1)^2 - 4(1)(m) = 4m^2+4m+1-4m = 4m^2+1 \]
Step 3: Solve $4m^2+1=0$.
\[ 4m^2 = -1 \implies m^2 = -\frac{1}{4} \] A square of a real number can never be negative. So this equation has no real solutions.
Step 4: Interpret what this means.
The discriminant $4m^2+1 \geq 1 > 0$ for all real $m$. This means the original quadratic ALWAYS has two distinct (unequal) real roots, no matter what real $m$ is chosen. It can never have equal roots.
Step 5: Count the number of valid $m$ values.
Since no real $m$ satisfies the equal-roots condition, the number of real values of $m$ is $0$.
Step 6: State the answer.
\[ \boxed{0} \]