When the parabola has only one intersection with the X axis (that is, δ = 0) and E is the vertex, E and F will coincide, then x 1=x2.

I have a simpler way to simplify.

Open the bracket on the left.

a+b-ax 1^2-bx 1

____________= 1-x 1

b-a

a( 1-x 1^2)+b( 1-x 1)

_ _ _ _ _ _ _ _ _ _ _ _ _ _ = 1-x 1

b-a

a( 1+x 1)( 1-x 1)+b( 1-x 1)

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 1-x 1

b-a

[a( 1+x 1)+b]( 1-x 1)

_ _ _ _ _ _ _ _ _ _ _ _ _ _ = 1-x 1

b-a

Divide both sides by 1-x 1 at the same time to get

a( 1+x 1)+b

________= 1

b-a

a( 1+x 1)+b=b-a

a( 1+x 1)+a=0

a(2+x 1)=0

Because a≠0

So 2+x 1=0.

So x 1=-2.