What are the solutions of the equation x^4+4x^3+2x^2-4x-m=0 and what is the value of m if equation has real solutions?

We'll re-write the equation into an equivalent
form:

x^2(x+2)^2 - 2x(x+2) - m =
0

We'll substitute the product x(x+2) by
t:

t^2 - 2t - m = 0

We'll
apply the quadratic formula:

t1 = [2 + sqrt(4 +
4m)]/2

t1 = [2 +
2sqrt(1+m)]/2

t1 = 1 +
sqrt(1+m)

t2 = 1 -
sqrt(1+m)

But t = x(x+2)

We'll
remove the brackets:

x^2 + 2x - t1 =
0

x^2 + 2x - 1 - sqrt(1+m) =
0

x1 = -1+sqrt[2+sqrt(1+m)]

x2
= -1-sqrt[2+sqrt(1+m)]

x^2 + 2x - t2 =
0

x^2 + 2x - 1 + sqrt(1+m) =
0

x3 = -1+sqrt[2-sqrt(1+m)]

x4
= -1-sqrt[2-sqrt(1+m)]

The roots x1,x2,x3 and x4 are real
if and only if the m satisfies the
constraints:

1+m>=0 =>
m>=-1

2 - sqrt(1+m) >=
0

m belongs to [-1 ; +infinite) and (-infinite ;
3]

For the equation to have all real
solutions, m = [-1 ; 3].