So given the cumulative distribution function (c.d.f.)
$P(r)=1-(2r^2+2r+1)e^{-2r}$ for $\: r \geq 0$
what would the median value be?
I tried setting the c.d.f. to $0.5$, which is the same as trying to solve $$\displaystyle \int \limits_{-\infty}^{R} p(r)\: \mathrm d r = 0.5,$$ where $p(r)$ is the probability distribution function, but that doesn't really help.
I'm going to assume that you meant to write the CDF as $$ P(r) = 1 - (2r^2 + 2 r + 1) e^{-2 r}\, . $$ (Note the minus sign in the exponent.) I also assume that the intended range of $r$ is $0$ to $+\infty$. This makes $P(r)$ satisfy $P(0) = 0$ and $\lim_{r\rightarrow +\infty} P(r) = 1$, as the CDF of a positive random variable should. The median value of $r$ satisfies $$ P(r) = 1 - (2r^2 + 2 r + 1) e^{-2 r} = 1/2\, . $$ Unfortunately, this equation has no nice analytic solution for $r$ so far as I'm aware. One can solve it numerically via various methods (I just used Mathematica) to find that the median value occurs at $r \approx 1.33703015686178$.