Unless the second degree polynomial has a double root then there is no simple formula for this. So the Taylor series are probably the best way to express your formula if you want to avoid the square root. This trades a square root for an infinite power series with limited region of convergence.

(-b+sqrt(b*b-4*a*c))/(2*a) and (-b-sqrt(b*b-4*a*c))/(2*a)

And if b*b-4*a*c

But what is the right hand side?

If the equation is sqrt(a*x*x+b*x+c) = 0, then its solutions are the same as for

a*x*x + b*x + c = 0, at least when the real domain is all you need. The situation isn't any different, when the solutions of the second equation are complex, say "z" and "z*" (I assume a,b, and c - real). I don't see any need for Taylor series.

Dear Marek Gutowski

There is no the right hand size:

sqrt(f(x))=f0(x)*(1+d(x)/2+...), where f(0) and d(x) are polynomials

Now I am puzzled. Now I see the right hand side, but what is the unknown? Seems that it is not "x" but rather many (?) functions: f0(x), d(x), ... - all being polynomials. If so then:

- arbitrarily good uniform approximation of a given sqrt{} is possible (on its domain, of course) with polynomial(s) (one of the numerous Cauchy theorems, if I remember well)

- but: 1) there is no universal approach how to do it (you may try any family of orthogonal polynomials on domain of sqrt{}), and 2) there is no guarantee that you ever obtain strict equality with finite number of terms.

I used orthogonal polynomials. The problem is the very high degrees of these polynomials for a "good" approximation. Thank you for your answer.

1. We have y(x)=sqrt(a*x*x+b*x+c) for x=[x1,x2]

2. One can approximate y(x) by the second order polynomial by MATLAB

3. Now it is very interesting to calculate the polynomial coefficients analytically from

a, b and c using constrains x=[x1, x2]. Numerical solutions are well known.