How to find the roots of s^3 + 6s^2 + 18s + 20 =0 ???

this equation s^3 + 6s^2 + 18s + 20 =0 is written as
(s+2)(s^2 + 4s +10 ) =0

how this 2nd equation is obtained from the 1st equation ????

Define the polynomial P(s) = s^3 + 6s^2 + 18s + 20
Try different values of s: when you get 0, that value is a root of the equation.
P(-1) = -1 + 6 - 18 + 20 = 7 (since you already know s = -2 is a solution or root you don't need to do this, but I write it as an example)
P(-2) = -8 + 24 - 36 + 20 = 0

So the polynomial is divisible by s+2. To factor it out, you can divide the normal way or use Ruffini's rule which is quicker:
.....| 1 6 18 20
.....|
..-2|...-2 -8 -10
-----|------------------------------
.....| 1 4 10 0

So (s+2)(s^2 + 4s + 10) = 0
One root is s = -2
Solve the second degree equation: s = (-4 +- sqrt[16 - 40]) / 2 = (-4 +- 2 sqrt[-1]) / 2 = -2 +- sqrt[-1]
There is no real solution. In imaginary numbers, let i = sqrt [-1]. Then the other two roots are
s = -2 + i
s = -2 - i

The Rational Roots Theorem gives us a clue on searching for rational roots :
(searching for roots in the form +- p/q with p=divisor of 20 and q=divisor of 1)
-1*2/1 is a root
There is a method to solve a cubic equation in general by hand (and calculator) on paper.
Substituting x=y+p in x^3+ax^2+bx+c yields :
y^3 + (3p+a) y^2 + (3p^2+2ap+b) y + p^3+ap^2+bp+c
if we take 3p+a=0 or p=-a/3, the first coefficient becomes zero, and we get :
y^3 + 6 y
(with p = -2)
Substituting y=qz in y^3 + b y + c = 0, yields :
z^3 + b z / q^2 + c / q^3 = 0
if we take q = sqrt(|b|/3), the coefficient of z becomes 3 or -3, and we get :
(here q = 1.41421356)
z^3 + 3 z + 0.00000000
Substituting z = t - 1/t, yields :
t^3 - 1/t^3 + 0.00000000
Substituting u = t^3, yields the quadratic equation :
u^2 + 0.00000000 u - 1 = 0
A root of this quadratic equation is u=1.00000000.