How to solve nonlinear equation with unknown variables?

I'm trying to solve the following equation using matlab 'syms'.

my code is:

syms x
solution = solve (W_c*(1 + (5/8)*(0.85*((2*x/delta_c - 1)^2 - 1) + 0.3*(2*x/delta_c - 1)^2*log(2*x/delta_c - 1))) - (1/2)*m*v_in^2 == 0, x);
delta_m = solution;

and my answer is:

delta_m =
 
  (6284008282358037*exp(lambertw(0, (262048699488897879269*exp(17/3))/25743973330415775)/2 - 17/6))/2417851639229258349412352 + 6284008282358037/2417851639229258349412352
 (6284008282358037*exp(lambertw(-1, (262048699488897879269*exp(17/3))/25743973330415775)/2 - 17/6))/2417851639229258349412352 + 6284008282358037/2417851639229258349412352
  (6284008282358037*exp(lambertw(1, (262048699488897879269*exp(17/3))/25743973330415775)/2 - 17/6))/2417851639229258349412352 + 6284008282358037/2417851639229258349412352

I don't know why it gives me such an answer, cause I suppose it should give me a real number.

I'm not sure if I'm using the correct code, can someone please help me with that?

Answer 1

[Walter Roberson]

No, it should not give you a real number. You did not tell MATLAB that x is restricted to reals, so it is proper that solve() returns all of the mathematically correct solutions it can find.

You can use

syms x real

This is not guaranteed to only return reals, so you might need to proceed to

x(imag(x) ~= 0) = [] ;

I also suspect that you will want to vpa(x) to see a numeric approximation to the exact solution that is returned.

Remember that if you do not want infinitely precise closed form solutions then solve() is probably the wrong function to call and vpasolve() is more likely to be what you want.

[Shipu HAN]

Thanks a lot, that works.