You have $\displaystyle L = 2C + \pi((D2+D1)/2) + ((D2+D1)^2/4C)$ and want to solve for C. The first thing I notice is that "4C" in the denominator of a fraction. Just because I don't like fractions, I would multiply both sides of that equation by 4C:
$\displaystyle 4CL= 8C^2+ 2\pi(D2+ D1)C+ (D2+ D2)^2$
We can write that as $\displaystyle AC^2+ BC+ D= 0$ with A= 8, $\displaystyle B= 2\pi(D2+ D1)- 4L$ and $\displaystyle D= (D2+ D1)^2$.
Now, by the quadratic formula, $\displaystyle C= \frac{-B\pm\sqrt{B^2- 4AD}}{2A}$.
Simplifying a bit: P = pi, A = D2+D1 ; equation becomes:
L = 2C + P(A/2) + A^2/(4C) : I'm assuming the "4C" you show should be "(4C)"
Multiply by 4C:
4CL = 8C^2 + 2CPA + A^2
Re-arrange:
8C^2 + 2CPA - 4CL + A^2 = 0
8C^2 + (2PA - 4L)C + A^2 = 0
That's a quadratic: a = 8, b = 2PA - 4L, c = A^2 : solve it!
EDIT: sorry Halls, me no see yours...
anyhoo.....mine's better!!