Consider the following equation:
$$x=\frac{(p+x(1-p))\cdot c}{1-(p+x(1-p))(1-c)}$$
Solution should be $$x=\frac{cp}{(1-c)(1-p)}$$
I tried to rewrite the first equation in such that:$$x^2[c(1-p )-(1-p)] + x[2cp - c + 1-p]-pc=0$$
Applying quadratic formula leads to something, since the term under the root cannot be simplified.
Is there another way to compute that?
$$x^2 (c+p-cp-1)-x (c+p-2 c p-1)-c p=0$$ Discriminant is $$D=(c+p-2 c p-1)^2-4cp (c+p-cp-1)=c^2+p^2+1+2 c p-2 c-2 p=(c+p-1)^2$$ so $$x=\frac{(c+p-2 c p-1)\pm(c+p-1)}{c+p-cp-1};\;c\ne 1\land p\ne 1$$ denominator $p-cp+c-1=p(1-c)-1(1-c)=(1-c)(p-1)$ $$x_1=\frac{c+p-2 c p-1-c-p+1}{2(1-c)(p-1)}=\frac{-cp}{(1-c)(p-1)}=\frac{cp}{(1-c)(1-p)}$$ and $$x_2=\frac{c+p-2 c p-1+c+p-1)}{2(1-c)(p-1)}=\frac{-2 (c-1) (p-1)}{2(1-c)(p-1)}=1$$ if $p=1\lor c=1$ then $x=1$