The parabola with equation y=ax^2+bx+c is graphed below:

We can solve this  for  "a"

-3  = a (-4 -2)^2  + 1

-3 = a ( -6)^2 + 1

-4  = 36a

a  = -4/36  = - 1/9

So....we have this form

y  = (-1/9( ( x - 2)^2  + 1       to find the zeroes....let y  = 0

0  = (-1/9)(x - 2)^2 + 1

-1  = (-1/9)(x - 2)^2       multiply through by -9

9 =  ( x - 2)^2        take both roots

±√9  = x - 2

±3  =  x - 2    add 2 to both sides

±3 + 2  =  x

So....the roots are

x =  3 + 2  = 5  = m         and   x  =  -3 + 2  =  -1  = n

So

m - n  =   5 - (-1)   =  6