Problem with Courant's 'Introduction to Calculus and analysis' I am confused from 'Another example is furnished by the function $f(x) = x^2$.' I'm sure the previous pages and having read the book previously are not necessary as my problem is with the manipulations and not concepts.
I'm able to follow the manipulations but 
the confusion is as to how he arrived at the result 
$\delta$ $= -|x_o| + $($\epsilon$ $+|x_o|^2)^{1/2}$
Thank you for your time

 A: $$\delta (2|x_0|+\delta)< \epsilon$$
$$\delta^2+2\delta|x_0|-\epsilon<0$$
By using the quadratic formula
$$\frac{-2|x_0|-\sqrt{4|x_0|^2+4\epsilon}}{2} < \delta < \frac{-2|x_0|+\sqrt{4|x_0|^2+4\epsilon}}{2}$$
Hence it suffices to choose $\delta=\frac{-2|x_0|+\sqrt{4|x_0|^2+4\epsilon}}{2}=-|x_0|+\sqrt{|x_0|^2+\epsilon}$
A: Note the inequality established already:
$$
|f(x) - f(x_0)| \le \delta (2 |x_0| + \delta)
$$
Now substitute the chosen $\delta=\sqrt{\epsilon + |x_0|^2} - |x_0|$ and it follows that:
$$
\begin{align}
|f(x) - f(x_0)| & \le \left(\sqrt{\epsilon + |x_0|^2} - |x_0|\right) \left(2 |x_0| + \sqrt{\epsilon + |x_0|^2} - |x_0|\right) \\
 & = \left(\sqrt{\epsilon + |x_0|^2} - |x_0|\right)\left(\sqrt{\epsilon + |x_0|^2} + |x_0|\right) \\
 & = \epsilon + |x_0|^2 - |x_0|^2 \\
 & = \epsilon
\end{align}
$$
A: 
I'm able to follow the manipulations but the confusion is as to how he arrived at the result

Judging from the text, at this point you aren't expected to be able to come up with that formula, or any other suitable formula, for $\delta$ yourself. Its appearance here is just for exposition: simply to show that a formula does exist to make it work.
You should, however, be able to plug in the formula into the definition of continuity to check that it really does show that $x \mapsto x^2$ is continuous at $x_0$.
Also, and this very important: the formula is not a result. Working with the $\epsilon$-$\delta$ definition of limits is not a "solve for $\delta$" type of problem. When a limit exists, there will be a whole interval of values for $\delta$ that suffice, and as long as you find anything in that interval, you've succeeded.
A: Alternative derivation:
You want to find a $\delta$ such that for a given $\epsilon$, $|x-x_0|<\delta\implies|x^2-x_0^2|<\epsilon$.
You can rewrite the condition as
$$x_0^2-\epsilon<x^2<x_0^2+\epsilon,$$
then assuming $x>0$,
$$\sqrt{x_0^2-\epsilon}-x_0<x-x_0<\sqrt{x_0^2+\epsilon}-x_0.$$
Taking the absolute value,
$$|x-x_0|<\min\left(\left|\sqrt{x_0^2-\epsilon}-x_0\right|,\left|\sqrt{x_0^2+\epsilon}-x_0\right|\right)=:\delta.$$

One may struggle to establish which of these bounds is the smallest (actually the "$+$" one by concavity of the square root). Anyway as none of them is zero, the "$\min$" expression is sufficient for the proof.
A: You start from 
$$
2|x-0|δ+δ^2<ϵ.
$$
To complete the square, add $|x_0|^2$ to both sides,
$$
(|x_0|+δ)^2<ϵ+|x_0|^2\iff -\sqrt{|x_0|^2+ϵ}<|x_0|+δ<\sqrt{|x_0|^2+ϵ}
$$
The left inequality is trivially true, the right gives the claimed bound
$$
δ<-|x_0|+\sqrt{|x_0|^2+ϵ}=\fracϵ{|x_0|+\sqrt{|x_0|^2+ϵ}}
$$
As all transformations were equivalences, the choice of such a $δ$ satisfies the continuity condition.

Note that the more general technique is to demand $δ=\min(1,\bar δ)$ so that one can extract one linear factor $δ$ and bound all $δ$ in the remaining factor to $1$,
$$
δ(2|x_0|+δ)\le \bar δ(2|x_0|+1).
$$
This will have a bound $ ϵ$ if $\bar δ=\frac{ϵ}{2|x_0|+1}$ is chosen. All in all
$$
δ=\min\left(1,\frac{ϵ}{2|x_0|+1}\right)
$$
is a valid choice.
A: As pointed out in the comment, in order for the continuity assumption to be satified, $|f(x)-f(y)|<\epsilon$ for which we may put $\delta(2|x_0+\delta)<\epsilon$, which is quadratic inequation, solving which we get the given value of $\delta$
