Derivation by completing the square method is incredibly simple, depending upon who is teaching the lesson... This is my very simple solution using algebra to show how the general formula is derived. This is as simple as it gets and I have chosen a different route to avoid messy fractions under the square root at the final stages.
We start by moving the constant term to the RHS.
We then divide throughout by a, thereby freeing up the x² term of its leading coefficient. In completing the square method, you always choose the b/a coefficient and divide it by 2 as shown below.
When you divide it by 2, it simply becomes b/2a, and there is nothing amazing going one here.
You then take b/2a term and square it as shown above. This is part of the “Completing the Square” mechanism.
You then add the result from the previous stage to both sides of the equation as shown above. By adding it to both sides of the equation, we have not changed anything and the equation is still balanced.
We now use the term b/2a that we calculated earlier to complete the square. By completing the square, the LHS term simplifies as shown above. The RHS stays the same as in the previous stage.
This might be a little controversial, but I prefer to multiply throughout by 4a² as you can see. By doing it this way, you can avoid many ugly fractions later.
As you can see, the RHS is starting to look familiar and very similar to the general form of the equation. On the LHS, we square the two terms. You need to bring them into a single squared bracket.
Simple algebra states that these two expressions are the same, so I am going to replace the 4a² term with (2a) ² as shown below. I am going to use this neat little mechanism to simplify the RHS.
Since the bracketed terms on the LHS are both squared, you can now bring the square outside to a single large bracket as shown below. This little rearrangement can save a lot of complication later on. I did this proof a couple of times to find the simplest way to show students.
Now that we have the square outside a big bracket, we can perform a square root on both sides to eliminate it.
As you can see, we now have a clean discriminant on the RHS. I call it clean because it is free of any messy fractions. I have seen many proofs where the “experts” have made a mess...
Now we divide both sides by 2a, to bring the 2a term under the radical. The radical is the square root by the way. I am calling it radical to make myself look clever... :-)
As you can see, it is already starting to look like the general form of the quadratic equation formula. I have arranged everything so that it starts to look familiar as early as possible.
Here we simply bring the b/2a term to the RHS to solve for x.
Because the denominator is the same, 2a, we can combine the numerator for both fractions onto a single common denominator as shown above to give us the famous expression.