Solve 4x^4 = 256x, using quadratic formula

4x^4 = 256x

Okay, I'm not completely inept, I've gotten part of it done:
(4x)x^3-64 = 0 --> 4x=(x=0)
(x-4)(x^2+4x+16)=0 --> x-4=(x=4)
-4+/-√16-64 / 2

So, two of the solutions are 4,0 and i know the next ones are going to be like -4-4i√4 or something but i'm so confused on where to go after you plug it into the quadratic formula. pleasee hellp. with detail, i have many more to go.

4x^4 - 256x = 0
4x(x³ - 64) = 0
4x(x-4)(x² + 4x + 16) = 0
From the zero product rule:
x = 0, 4
x²+4x+16 = 0
x = [-4±√(16-64)]/2 = -2 ± √-12 = -2±2i√3

The four roots of the equation are:
x = 0, 4, -2-2i√3, and -2+2i√3

4x^4 = 256x
x^4-64x=0
x(x^3-64)=0
x^3-64=0
x = 4
x = 4 or 0
Sorry I misunderstood your question I thought you were trying to solve the original equation using the quadratic formula.
Once you have the (x^2+4x+16) factor you could avoid the quadratic formula by completing the square.
(x+2)^2 = -12
x = -2 ±2i*sqrt(3)

u dont need quadratic formula for this question.
First divide 4 from both sides.
4x^4 = 256x
x^4 = 64x
x^3 = 64
x = 4

(x^2+4x+16)=0
▲= b²-4ac
▲= 16-4*16
since ▲<0, √▲ yields two complex numbers
x= (-b±√▲)/2a