Finding true roots of $x^2 + 10^9x +1$ using the Quadratic Formula I've been given a hint to rearrange the quadratic formula to:
$$b\sqrt{1-\frac{4ac}{b^2}}$$
but am still dealing with a numerical error at $4/10^{18}$
Can I solve this numerically with the Taylor Series?

Almost as H. H. Rugh answered, 
consider the case of $$x^2+a x+1=0$$ where $a$ is a huge number.
The roots are given by $$x_{1,2}=-\frac{1}{2} \left(a\pm\sqrt{a^2-4}\right)=-\frac{a}{2} \left(1\pm\sqrt{1-\frac 4{a^2}}\right)$$ Now, using Taylor series for small $y$ (or the generalized binomial expansion)$$\sqrt{1-y}=1-\frac{y}{2}-\frac{y^2}{8}-\frac{y^3}{16}-\frac{5 y^4}{128}+O\left(y^5\right)$$ replace $y$ by $\frac 4{a^2}$ to get $$x_1=-a+\frac{1}{a}+\frac{1}{a^3}+\frac{2}{a^5}+O\left(\frac{1}{a^6}\right)$$ $$x_2=-\frac{1}{a}-\frac{1}{a^3}-\frac{2}{a^5}+O\left(\frac{1}{a^6}\right)$$
