Solve the quadratic equation

(x-1)²=k²(x+k)
x^2-2x+1=k^2x+k^3
x^2-(2+k^2)x+1-k^3=0
x^2-(2+k^2)x+(1-k)(1+k+k^2)=0
(x-(1-k))(x-(1+k+k^2))=0
x=1-k or 1+k+k^2

(x-1)²=k²(x+k)
x^2-2x+1=(k^2)x+k^3
x^2-(2+k^2)x+(1-k^3)=0
Using the quadratic formula x=[-b+-{b^2-4ac}]/2a for the quadratic equation ax^2+bx+c=0
where {}is square root
So, x=[(2+k^2)+{(2+k^2)^2-4(1-k^3)}]/2 or x=[(2+k^2)-{(2+k^2)^2-4(1-k^3)}]/2
=[2+k^2+{4+4k^2+k^4-4+4k^3}]/2 or =[2+k^2-{4+4k^2+k^4-4+4k^3}]/2
=[2+k^2+{4k^2+4k^3+k^4}]/2 or =[2+k^2-{4k^2+4k^3+k^4}]/2
=[2+k^2+{k^2(4+4k+k^2)}]/2 or =[2+k^2-{k^2(4+4k+k^2)}]/2
=[2+k^2+{(k^2)(2+k)^2}]/2 or =[2+k^2+{k^2(4+4k+k^2)}]/2
=[2+k^2+k(2+k)]/2 or =[2+k^2-k(2+k)]/2
=[2+2k+2k^2]/2 or =[2-2k]/2
=1+k+k^2 or =1-k

So, The roots of (x-1)²=k²(x+k) are 1+k+k^2 and 1-k
i.e. x=1+k+k^2 or x=1-k