p(x)=2x^4-2x^2-12

OK......so we have

2x^4-2x^2-12 = 0   divide everything by 2

x^4 - x^2 - 6 = 0    we can factor this directly as

(x^2 + 2 ) (x^2 - 3) = 0

Setting the first two factor to 0, we have that x^2 + 2 = 0  ... subtract 2 from both sides....

x^2 = -2      take the square root of both sides

x = ±√(-2) = ±i√(2) or ±√(2)i, if you prefer  .....and those are the complex roots.....

And setting  x^2 - 3 = 0 and adding 3 to both sides, we get

x^2 = 3     take the square root of both sides and we get......

x = ±√(3)  ....and those are the real roots....

What's the question???

I'm trying to find all real and/or complex zeroes of p(x)