How to write code to find the square root of Integer
Find a square root of Integer, it means to solve a polynomial $f(x) = x^2 - n = 0$ \begin{align*} f(x) &= x^2 -n \\ 0 &= x^2 -n \\ n &= x^2 \\ \sqrt{n} &= x \\ \end{align*} The above equations DO NOT really tell you how to find the square root of $n$
Let assume me try to find the square root of 4, then we have: \begin{align*} x^2 - 4 &= 0 \\ \end{align*} The graph of $x^2 - 4 = 0$ is below

From the picture above, we can guess a number on x-Axis right hand side of $\color{red}E$ e.g. choose point $C$ as on the picture.
and find the middle point $m$ from $A$ to $C$,
\[ m = m' = \frac{A + C}{2} \\ \Rightarrow y = m^2 - 4 \\ \] If $y < 0$ then $m$ will be left hand side of true value $\color{red}E$, otherwise $m$ will be right hand side of true value $\color{red}E$.
If $m^2 < 0$ then the true value of $\sqrt{n}$ must be between $m$ and $C$
If $m'^2 \ge 0 $ then the true value of $\sqrt{n}$ must be between $A$ and $m'$


Let write some code in Swift
Note: 1. we want to keep the approximate value close to 
         the true value not less than epsilon,[0.00001]

      2. If n = 1 then we can just return 1

func squareRoot(n:Int)->Double{
    // x^2 - n = 0 
    var a:Double = 1
    var dn = Double(n) 
    var b:Double = Double(n) 
    var m:Double = (a + b)/2.0
    // initial 
    let epsilon = 0.00001
    if n == 1{
        return 1
    }else{
        while m < epsilon{
            m = (a + b)/2.0
            if m*m - dn < 0{
                a = m
            }else{
                b = m
            }
        }
    }
    return m
}
Other method to find the square root of n is Newton method.

1. Choose any point $(x_0, f(x_0))$ on the curve
2. Compute the epsilon $\varepsilon = \left|x_0^2 - n \right|$
3. If $\varepsilon$ satified the boundary condition: $\left|x_0^2 - n \right| < \varepsilon$, e.g. $\varepsilon = 0.001$
$\quad$ 3.1 Terminate the code
4. Otherwise, find the tangent line at the point $(x_0, f(x_0))$
5. Find the intersection $(x_1, 0)$ of tangent line and x-Axis
6. Go back to 1. and use point $x_1$ instead of $x_0$ to continue

Here is the derivation on Newton's method
\begin{align*} f'(x_0) &= \frac{y_1 - y_0}{x_1 - x_0} \\ x_1 - x_0 &= \frac{y_1 - y_0}{f'(x_0)} \\ x_1 &= x_0 - \frac{y_0}{f'(x_0)} \\ \\ x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)} \quad \text{ where } \quad f(x_0) = y_0 \\ \\ \text{ If } f(x) &= x^2 - n \\ x_1 &= x_0 - \frac{f(x_0)}{2x_0} \quad \text{ where } \quad f'(x_0) = 2x_0 \\ \end{align*}
Line tangents to $f(x) = x^2 - n$ at point $(x_0, f(x_0))$ is $f'(x) = \frac{y - y_0}{x - x_0}$ 

// In C++
double newton_method(int n){
    double x0 = 1;
    double x1 = 1;
    double epsilon = 0.001;
    double delta = abs(x0*x0 - n);
    while (delta > epsilon){ 
        x1 = x0 - (x0*x0 - n)/(2.0*x0);
        delta = abs(x1*x1 - n);
        x0 = x1;
    }
    return x1;
}

How to find the root of polynomial equations

Find the root of quadratic equation is pretty easy since we learn a quadratic formula in high school.

\begin{align*} &y = ax^2 + bx + c = 0 \\ &x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} \\ \end{align*} Let derive the formula \begin{align*} &y = ax^2 + bx + c \\ &y = x^2 + \frac{b}{a} x + \frac{c}{a} \\ &y = x^2 + 2 \frac{b}{2a} x + \frac{c}{a} \\ &y = x^2 + 2 \frac{b}{2a} x + (\frac{b}{2a})^{2} - (\frac{b}{2a})^{2} + \frac{c}{a} \\ &0 = (x + \frac{b}{2a})^2 - (\frac{b}{2a})^{2} + \frac{c}{a} \quad \text{ where } y = 0 \\ &(x + \frac{b}{2a})^2 = (\frac{b}{2a})^{2} - \frac{c}{a} \\ &(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \\ &(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \\ &x + \frac{b}{2a} = \frac{\pm \sqrt{b^2 - 4ac}}{2a} \\ &x = \frac{\pm \sqrt{b^2 - 4ac}}{2a} - \frac{b}{2a} \\ &x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} \\ \end{align*} Find the root for quadratic equation is easy because we have formula for it.

How to find [all] root of any equations


Technically, it is hard problem in mathematics. But if we can restrict ourself in certain interval,
it might be easier to find all the roots from them.
Actually, there are formula to compuate the upper and lower boundary for all the real roots in any degree of polynomial. \begin{align*} y &= (x-1)(x-4)x \end{align*} First, we look at the a single interval $[A, C]$ which contais $E$


Second, we divide the large interval: $[-2, 6]$ to small intervals: $[F, G) [G, H) \dots [E, C]$


Third, we can compute all the small intervals to find all the roots in $[-2, 6]$


Algorithm:
0. Given interval: \[ [a_0, a_n] \] 1. Divide a given interval to $n$ smaller intervals
\[ \delta = \frac{a_n - a_0}{n} \\ \text{ where } a_0 = \delta_0 \quad a_n = \delta_n \\ \alpha_1 = [\delta_0, \delta_1) \\ \alpha_2 = (\delta_1, \delta_2] \\ \vdots \\ \alpha_n = (\delta_{n-1}, \delta_n] \\ \] 2 Iterate each interval:
If one root is found,
then Add the root to list
Go back to 2