Question

How do you solve for x in `""^3sqrt(-2-5x)+3=0`?

Answer 1

I tried this:

Explanation:

I would try to rearrange and apply the power of `3` to both sides:
`root3(-2-5x)=-3`
raise to power `3`:
`(root3(-2-5x))^3=(-3)^3`
`-2-5x=-27`
`-5x=2-27`
`5x=25`
`x=25/5=5`