Find x for (x^2 - 16)^2 + 2x^2 – 28 = 0

Find x for (x^2 - 16)^2 + 2x^2 – 28 =
0

Put x^2 -16 = t.Then the  equation
becomes:

Then 2x^2 =
2(t+16).

So the given equation
becomes:

t^2+2(t+16)-28 =
0

t^2+2t +4 = 0

t =
{-2+or-sqrt(2^2-4*1*4)}/2 , Or

t1 = {-1 +sqrt (-3)}. Or t2
=
{-1-sqrt(-3)}

Therefore ,

 x^2-16
= -1 +sqrt(-3) gives x = 16-1 +sqrt(-3(

x^2 = 15+sqrt(-3) ,
or x^2 = 15-sqrt(-3).

Therefore there are 4 complex roots
which are below:

x1 = sqrt{15+sqrt(-3)} or x2=
sqr{15-sqrt(-3)}

x3 = -sqrt{15-sqrt(-3)} or x4 =
-sqrt{15-sqrt(-3)}