Derivaiton Of The Quadratic Formula

First you divide both sides by "A" so you get x^2+(Bx/A)+(C/A)=0

Then take "(C/A)" to the other side so you get x^2+(Bx/A)=-C/A

Add "(B/2A)^2" to both sides, so you get

(x^2)+(Bx/A)+(B/2A)^2=-C/A +(B/2A)^2

We can rewrite this as

[x+(B/2A)]^2=-C/A +(B/2A)^2

Now solve for x, we start with square rooting everything, so you get

x+(B/2A)=[-C/A +(B/2A)^2]^0.5 x+(B/2A)=-[-C/A +(B/2A)^2]^0.5 (something to the power of 0.5 is square rooted)

Then we move "(B/2A)" to the other side which gets us

x=-(B/2A)+[-C/A +(B/2A)^2]^0.5 x=-(B/2A)-[-C/A +(B/2A)^2]^0.5 This is solved but we can simplify it, let`s multiply the right by 2A/2A

x=-B+{2A/2A[-C/A +(B/2A)^2]}^0.5 x=-B-{2A/2A[-C/A +(B/2A)^2]}^0.5

Lastly, we can simplify this to get the quadratic formula we all know and love, which is

x={-B+[(B^2)-4AC]^0.5}/2A x={-B-[(B^2)-4AC]^0.5}/2A

Sorry for the bad layout, you can`t really write it well on this, but I hope you learnt something