Question

How do you solve `3x^4+12x^2-15=0`?

Answer 1

Assuming we are restricted to Real (non-Complex) solutions:
`color(white)("XXX")x=+-1`

Explanation:

Given
`color(white)("XXX")3x^4+12x^2-15=0`

Dividing both sides by `3`
`color(white)("XXX")x^4+4x^2-5=0`

Temporarily substituting `q` for `x^2`
`color(white)("XXX")q^2+4q-5=0`

Factoring the left side
`color(white)("XXX")(q+5)(q-1) = 0`

With potential solutions
`color(white)("XXX")q=-5` and `q=1`

If `q=-5`
then
`color(white)("XXX")x^2=-5`
which is impossible for Real values of `x`

If `q=1`
then
`color(white)("XXX")x^2=1`

`rarrcolor(white)("X")x=+-1`