Nonlinear equation / iteration method $$ 1 x^5-9.067 x^4+24.726 x^3-13.998 x^2-15.278 x+1.014 = 0 $$
Wolfram Result
but i need solve roots using "iteration method" (i don't know how it correctly called in English, but Newton's method is not what I need)
method, what I need, starts from transformation of equation
fi(x) = x; if |fi'(x)| < 1 we can find root
I used fi(x) = - (1 x^5 - 9.067 x^4 + 24.726 x^3 - 13.998 x^2 + 1.014) / 15.278
but with it I solved just two roots from five
 A: There are a number of iterative techniques that can be used to solve a equations.  Here's one:  If you write this equation in the form
$$x = (-1.014 + 13.998 x^2 - 24.726 x^3 + 9.067 x^4 - x^5)/15.278,$$
Then you see that any root of your polynomial is a fixed point of 
$$f(x) = (-1.014 + 13.998 x^2 - 24.726 x^3 + 9.067 x^4 - x^5)/15.278.$$
Thus, we might be able to solve your equation by iterating $f$.  In fact, the first seven iterates of $f$ starting at zero are:
$$0, 0.066369, 0.062795, 0.0631486, 0.063114, 0.063117, 0.063117, 0.063117$$
In general the derivate at the fixed point should be less than one in absolute value for this to work.  I believe that there's one other solution that this technique can find.
If, as you say in your comment, you hope to find other roots, then you might try writing the equation in the form $x=f(x)$ in other ways.  One other way, for example is as
$$x = \sqrt[3]{\frac{-1000 x^5+9067 x^4+13998 x^2+15278 x-1014}{24726}}.$$
I obtained this by solving for just the "$x$" under the cube.  If we iterate this new function, we might converge to different roots.  The obit of zero, for example, is now
$$0, -0.34485, -0.56606, -0.55328, -0.55733, -0.55610, -0.55648, -0.55637$$
