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y = 3x^2  - 5x  - 3

Let   A  = (u, 3u^2 - 5u  -3)

Let B =   (v, 3v^2 - 5v - 3)

Using the midpoint, rule, we have that

[ u + v ] / 2  = 0         and      [ 3u^2 - 5u  - 3 + 3v^2 - 5v  - 3 ] / 2  = 0    (2)

u + v  = 0 

v = -u    (1)

Sub (1)  inot (2)  and we have that

[3u^2 - 5u - 3 + 3 (-u)2 - 5(-u) - 3 ] /  2   =  0            multiply  through by 2  and simplify

6u^2  - 6   = 0      divide through by 6

u^2  - 1   = 0     factor

(u + 1) ( u -1)  = 0

Set both factors to 0  and solve for  u  ⇒    u = -1    or  u  = 1

If  u   = -1    then the associated  y  coordinate is  3(1)^2 - 5(-1) - 3   = 5

So A  = (-1, 5)

And the x  coordinate  of  B  is  -u =  1  and  the y coordinate of  B is 3(1)^2 - 5(1) - 3 = -5

So, in this case B  = (1, -5)

The  slope of this line through A and B is    [ -5 - 5]  / [ 1 - -1] =  -10/2  = -5

And the equation of this line is

y  = -5 (x -1) - 5

y = -5x 

So....the  distance between  A  and B   is  √ [ -5 - 5)^2  + ( -1 -1)^2 ]  = √[ 10^2 + 2^2] = √104  

So....the square of the length of AB is just 104

Note that if u  = 1, we have the same results...A and  B are  just "switched"

See the graph, here :  https://www.desmos.com/calculator/e2uvdwwhl3