Solution:

Order √( 1+c)=k,

So 1+c=k? ，c=k？ - 1

k？ *( 1+k)=3*(k？ - 1)*k

k *( 1+k)(k-3 *(k- 1))= 0

k*( 1+k)(3-2k)=0

K=0 or k=- 1, k=3/2.

When k=0, c=- 1.

C=i- 1 (complex number) when k=- 1.

When k=3/2, c=5/4.

After testing, c=- 1 and c=5/4 (complex number c=i- 1) are all solutions of the original equation.