How do you factor completely $ {x^2} = 20 $ ?

Hint: In order to determine the value of x , by taking square root on both sides include $ \pm $ in the result.
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
 $ x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} $ and $ x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} $
 $ x_1 $ , $ x_2 $ are root to quadratic equation $ a{x^2} + bx + c $
 Hence the factors will be $ (x - x_1)\,and\,(x - x_2)\, $ .

Complete step-by-step answer:
Given a equation $ {x^2} = 20 $
 $
   \Rightarrow {x^2} = 20 \\
   \Rightarrow x = \pm \sqrt {20} - - - - - - (1) \;
  $
So to determine value of $ \sqrt {20} $
Separating the value $ 20 $ into its factors, So the factors of $ 20 $ comes to be,
 $ 1,2,4,5,10,20 $
Now let’s find the factors who are perfect squares of some number, we got
 $ 1,4 $
Let’s consider the largest perfect square form the factors of $ 20 $ and divide it with $ 20 $ ,we get
\[
   = \dfrac{{20}}{4} \\
   = 5 \;
 \]
From the above we can say that $ 20 = 4 \times 5 $
Replace $ 20 $ as $ 4 \times 5 $ in the original number
\[
  \sqrt {20} = \sqrt {4 \times 5} \\
   = \sqrt {{2^2} \times 5} \;
 \]
Taking out $ 2 $ from inside the square root.
 $ = 2\sqrt 5 $
Putting the value of $ \sqrt {20} $ in equation (1)
 $ \Rightarrow x = \pm 2\sqrt 5 $
 $ \Rightarrow x = + 2\sqrt 5 \,or\,x = - 2\sqrt 5 $
Therefore, value of $ x = + 2\sqrt 5 \,or\,x = - 2\sqrt 5 $
So, the correct answer is “ $ x = + 2\sqrt 5 \,or\,x = - 2\sqrt 5 $ ”.

Note: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions