Find a matrix P such that [x]r = P[x]b

stuck on this problem for my linear algebra homework. B is the basis consisting of the vectors
[5 , 0] and [-2 , 5] . These are both supposed to be vertical with the 5 and -2 on top, respectively. R is the basis consisting of [-2 , 3] and [1 , -2], and these are also vertical.
I figure that P would just be [x]r times the inverse of [x]b , but this is not the right answer. Thanks for any help you can provide! I also do pick best answer, so you can rest in peace about that haha.

Let R and B represent the matrices

[-2 1]
[3 -2]

and

[5 -2]
[0 5]

respectively.

Then for any column vector x in R^2,
[x]r = R^(-1) [x] = R^(-1) (B[x]b) = (R^(-1) B)[x]b.

So if we choose P = R^(-1) B, then [x]r = P[x]b.

P = R^(-1) B

=
[-2/detR -1/detR][5 -2]
[-3/detR -2/detR][0 5]

=
[-2/(4-3) -1/(4-3)][5 -2]
[-3/(4-3) -2/(4-3)][0 5]

=
[-2 -1][5 -2]
[-3 -2][0 5]

=
[-10 -1]
[-15 -4].

Lord bless you today!