Solving for x Intercepts to Polynomial Functions

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I'm having trouble finding x intercepts i have the question x^3-9x^2+15x+30 and by setting y=0 i got -30=x(x^2-9x+15) I used the quadratic formula to find that x^2-9x+15 give me no real roots, so the only intercept would be x=0. But this would also mean that the yintercept is 0 and i found that it is 30. What am i doing wrong is it something to do with moving the 30 to the left side?
 
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-30=x(x^2-9x+15)

You can not solve the right hand side to be = 0 if you have left hand side = -30 ...
 
so how do i do it. i was moving that over to solve for x
 
test solution, then rewrite x^3-9x^2+15x+30 as (x+n)(x-b)(x-4) or something in that style.

If c solves the p(x) = then p(x) can be divided with (x-c)..
and vice versa, (x+r) ; P(-r) = 0
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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