Solve Spherical Shell for Radius given Thickness and Volume

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I'm looking to calculate the outer radius of a spherical shell of a desired volume and thickness. I don't know if the years have knocked some obvious obstacle out of my perception, but here's what i'm struggling with.
Given: Thickness T, Radii R and r, and Volume V and the following relationships: $$ T=R-r $$ $$ V=\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 $$ How do I solve for R in terms of V and T?

I get as far as:

$$ \frac{3V}{4\pi}=R^3-(R-T)^3 $$ Which comes down to... $$ \frac{3V}{4\pi} = 3R^2-3RT+T^2 $$

Which i'm lost on breaking apart. For more clarification, this is supposed to be just a fun exercise for a D&D game. The rules i'm working with are limited by volume, and the integrity of created objects would be limited by thickness. Thus, these are the terms i wanted to control, and then calculate the resulting outer radius of what the character can create.

Thanks for any help that can be provided!

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This is the right way to do it (except you have an algebraic error in your last equation, which you could deduce from dimensional analysis). Now you have a simple quadratic in $R$: $$R^2 - TR + \left(\frac{T^2}{3} - \frac{V}{4\pi T}\right) = 0.$$

Use the quadratic formula to solve it, and pick the positive root.

$$R = \frac{T + \sqrt{T^2 - 4\left(\frac{T^2}{3} - \frac{V}{4\pi T}\right)}}{2} = \frac{T + \sqrt{T^2 - \frac{4}{3}T^2 + \frac{V}{\pi T}}}{2} = \frac{T+\sqrt{\frac{V}{\pi T} -\frac{T^2}{3} }}{2}$$