How to find zeros for 4x^4-11x^2-3

I dont understand.

4x^4 - 11x^2 - 3

Finding zeros means setting this equal to 0 and solving for x:

4x^4 - 11x^2 - 3 = 0

Let u = x^2:

4u^2 - 11u - 3 = 0

Now, factor as you would any quadratic. Split the b-term:

4u^2 - 12u + u - 3 = 0

(4u^2 - 12u) + (u - 3) = 0

Factor the first term:

4u * (u - 3) + (u - 3) = 0

Re-combine:

(4u + 1) * (u - 3) = 0

So, we have:

4u + 1 = 0 and u - 3 = 0

This means:

4x^2 + 1 = 0 and x^2 - 3 = 0

Solve for x:

4x^2 = -1 and x^2 = 3

x^2 = -1/4 and x^2 = 3

x = i/2 and i/2 and sqrt(3) and -sqrt(3)