What are the zeroes of the equation?

the best placed to go for questions like these is www.wolframalpha.com.The answer I got was -5, 5.
What I did was factor the problem.
-3x^4+27x^2+1200=0
Factor out -3
-3[x^4-9x^2-400]
Factor the inside(16 & 25 are the factors)
-3(x^2+16)(x^2-25)
Factor the far right-Rule(A^2-B^2)=(A-B)(A+B)
-3(x^2+16)(x+5)(x-5)=0
and there you go. I did not neglect the x^2+16 because x^2=-16 is imaginary.

0 = -3x^4 + 27x^2 + 1200

= x^4 - 9x^2 - 400

= (x^2-25)(x^2+16)

=> Reject x^2 + 16 > 0

=> x^2 = 25

=> x = ± 5

x=+/-5 ,+/-4i

-3x^4+27x^2+1200=0 , div by -3 =>
x^4 - 9x² - 400 = 0

x² = 9/2 +/- √(81/4+1600/4)
x² = 9/2 +/- 41/2

x² = 50/2 = 25 --> x = +/-5
x² = -32/2 = -16 --> x = +/- 4i

-3x⁴ + 27x² + 1200 = 0
x⁴ - 9x² - 400 = 0

Quadratic formula
x² = [9 ± √(9² – 4·1(-400))] / [2·1]
 = [9 ± √1681] / 2
 = [9 ± 41] /2
 = -16, 25
√(-16) = 4i
√25 = 5

x = ±4i, ±5

x = 5 or x = -5 or x = 4i or x = -4i