Basic algebra for solving an equation, or not?

Question

I have a "simple" equation which I have to solve for $ R_2 $: $$ \arccos\left(\frac{\sin\left(\arctan\left(\frac{M}{R_1+R_2-M}\right)\right)}{\sin\left(\frac{\pi}{T}\right)}\right)=\arctan\left(\frac{S\sqrt{R_1^2+R_2^2}}{2\pi R_2}\right) $$ Where: $$ R_1,R_2,M \in \mathbb{R}_{>0} : \left\{ R_1,R_2,M \in \mathbb{R} \mid R_1,R_2,M > 0 ; R_2<R_1 \text{ and } M\le\frac{R_1-R_2}{2}\right\}$$ $$ T \in \mathbb{N}_{>2} : \left\{ T \in \mathbb{N} \mid T > 2 \right\}$$ $$ S \in \mathbb{Q}_{>0} : \left\{ S \in \mathbb{Q} \mid S > 0 \right\}$$ $$ $$ I've managed to eliminate all the trigonometry using $\cos(\arctan(x))$ formula instead of $\tan(\arccos(x))$ which I've found to be a little more complicated and I'm left with a monstrosity of a quartic equation $$ \Bigl(4\pi^2y\Bigr)R_2^4 + \Bigl(8\pi^2My\Bigr)R_2^3 +\Bigl(4\pi^2yR_1^2+8\pi^2M^2y-8\pi^2R_1My-4\pi^2M^2+M^2S^2\Bigr)R_2^2 - M^2S^2R_1^2=0 $$ where $ y=\sin^2(\frac{\pi}{T}) $. This perhaps has complex roots which I definitely don't want: I know with certainty the equation has to have a real root for $R_2$. How can I find it ?