The quadratic formula is x = [-b ± √(b2 - 4ac)]/2a (1) ,where a,b and c are the coefficents of a quadratic in the form ax^2+bx+c=0 (2).
To prove that this works we need to start from equation (2)and using algebra get to equation (1).
So our first step will be to take equation (2) and divide it all by a. This leaves us with x^2+(b/a)x+(c/a)=0 (3). We are allowed to do this as we know that a is non zero.
Then we subtract (c/a)from both sides of equation (3), which results in x^2+(b/a)x=-(c/a) (4).
In the next step we want to complete the square so we have to add (b/2a)^2 to both sides of equation (4). This the gives us x^2+(b/a)x+(b/2a)^2=-(c/a)+(b/2a)^2 (5).
We then complete the square on the LHS of (5) to give us [x+(b/2a)]^2=-(c/a)+(b/2a)^2 and here the RHS can als be re-written to give us (b^2-4ac)/2a to give us a full equation of [x+(b/2a)]^2=(b^2-4ac)/2a (6).
Now we can take the square root of both sides of (6) to give us x+(b/2a)=(sqrt(b^2-4ac)/2a) (7).
Now all we have to do is minus (b/2a) from both sides, this results in the quatratic formula of x = [-b ± √(b2 - 4ac)]/2a (1) so there we have proven why the quadratic formula works.