Find all the series of the equation.

I meant zeroes, not series

what ypu have is a quadratic in x^2:

-3(x^2)^2 + 27(x^2) + 1200 = 0
so, using the quadratic formula,

x^2 = (-27±√15129)/-6 = (-27±123)/-6
x^2 = -16 or 25
so, x = ±4i,±5

First< I would divide each term by -3 to get

x^4 - 9x^2 - 400
can you think of 2 numbers which multiply to get -400, and add to -9 ?
how about -25 and 9 ?
(x^2 + 9)(x^2 - 25) = 0

x^2 = -9 or x^2 = 25
x = ± 3i or x = ± 5

oops, first factor is x^2 - 16

so x = ± 4i, see oobleck's above

To find the series of the given equation, we can use the quadratic formula. The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions can be obtained using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our given equation: -3x^4 + 27x^2 + 1200 = 0, we can rearrange it to match the quadratic form by dividing through by -3:

x^4 - 9x^2 - 400 = 0

Comparing this equation with the quadratic form ax^2 + bx + c = 0, we find that a = 1, b = 0, and c = -400.

Now, let's substitute these values into the quadratic formula:

x = (-0 ± √(0^2 - 4(1)(-400))) / (2(1))

x = (± √1600) / 2

x = ± 40 / 2

Simplifying further, we have:

x = ± 20

So, the series of the equation is: x = -20 and x = 20.

By substituting these values back into the original equation, we can confirm that they satisfy it.