I let $x=\frac dR$ in Alpha and just plotted it. It gave me the approximate root $0.45125$ Given that, the $x^3$ term is rather small. I then wrote an iterative solution: $x_{i+1}=\frac 49(1+\frac {3x^3}{16})$ It converges rapidly to the correct answer. After five cycles starting with $0.4$ or $0.5$, it has converged to six places. Of course, Newton's would converge even faster, but this was easy to put together in Excel.

I let $x=\frac dR$ in Alpha and just plotted it. It gave me the approximate root $0.45125$ Given that, the $x^3$ term is rather small. I then wrote an iterative solution: $x_{i+1}=\frac 49(1+\frac {3x^3}{16})$ It converges rapidly to the correct answer. After five cycles starting with $0.4$ or $0.5$, it has converged to six places. Of course, Newton's would converge even faster, but this was easy to put together in Excel.

Added per request: $$\begin {array} 0.4&0.449777778\\ 0.449777778&0.45202695\\0.45202695&0.452141272\\0.452141272&0.452147113\\ 0.452147113&0.452147411\\0.452147411&0.452147427 \end {array}$$

To make such a spreadsheet, the first column is $x$, the second is $f(x)$ You put the starting value in the first cell of the first column, then each cell in the first column is the value in the cell in the second column in the row above. Copy down and you are done.