Help how would I solve this equation: z^4 = 4z^2 + 32

z^4 = 4z^2 + 32; subtract 4z^2 from both sides
z^4 - 4z^2 = 4z^2 - 4z^2 + 32; simplify
z^4 - 4z^2 = 32; subtract 32 from both sides
z^4 - 4z^2 - 32 = 32 - 32; simplify
z^4 - 4z^2 - 32 = 0; factor
(z^2 + 4)(z^2 - 8) = 0; set each factor equal to 0

z^2 + 4 = 0; subtract 4 from both sides
z^2 + 4 - 4 = 0 - 4; simplify
z^2 = -4; square root each side
√(z^2) = √(-4); simplify
z = +/-2i

z^2 - 8 = 0; add 8 to each side
z^2 - 8 + 8 = 0 + 8; simplify
z^2 = 8; take the square root of each side
√(z^2) = √(8); simplify
z = +/-√(8); reduce to simplest terms
z = +/-√(4)*√(2); simplify
z = +/-2√(2)

Blessings

z⁴ = 4z² + 32
z⁴ - 4z² - 32 = 0

Use the quadratic formula to solve for z²
z² = [-(-4) ± √((-4)² - 4(1)(-32))] / (2·1)
   = [4 ± √(144)] / 2
   = [4 ± 12] / 2
   = -4, 8

z² = 8 ⇒ z = 2±√2
z² = -4 ⇒ z = ±2i

quadratic equation applied twice first time set y=z^2 changing the equation to y^2=4y+32 then after you use the quadratic equation you will get two answers that you put z^2 back in for y and solve both then check your answers in the original

z^4 - 4z^2 - 32 = 0

=> z^4 - 8z^2 + 4z^2 - 32 = 0

=> z^2( z^2 - 8) + 4( z^2 - 8) = 0

=> ( z^2 - 8)(z^ + 4) = 0

=> z^2 = 8 , -4

=> z = +/- 2sqrt2 or z = +/- 2i

I go by DW answer. though both of the above are same but DW has made it seem simple..