If $(1-p)$ is a root of quadratic equation $x^2+px+(1-p)=0,$ then its roots are

Since $(1-p)$ is a root of $x^2+px+(1-p)=0,$ it must satisfy the equation:
$(1-p{)}^2+p(1-p)+(1-p)=0$.
$(1-p)[(1-p)+p+1]=0$.
$(1-p)[2]=0\Rightarrow 1-p=0\Rightarrow p=1$.
Substituting $p=1$ back into the original quadratic equation:
$x^2+(1)x+(1-1)=0$
$x^2+x=0$
$x(x+1)=0$
The roots are $x=0$ and $x=-1$.