-2X？ +6X-4=0

x？ -3X+2=0

(X- 1)(X-2)=0

X 1= 1，X2=2，

When X= 1 or 2, Y 1=Y2.

(2) Because the opening of the quadratic function is downward, the function value is greater than Y=-3X+6 between X= 1 and X=2.

That is, when 1 < x < 2 and y 1 < y2.

When x < 1 or x > 2, y 1 > y2.

(3) Substitute Y=0 into Y=-2X? +3X+2

-2X？ +3X+2=0，

2X？ -3X-2=0

(2X+ 1)(X-2)=0

X 1=- 1/2，X2=2

The coordinates of point B are (-1/2,0), and the coordinates of point C are (2,0).

The symmetry axis of the function is X=-3/(-4)=3/4, and the vertex is on the symmetry axis, so the abscissa of the vertex is 3/4.

Substitute x = 3/4, y = 25/8, and the coordinate of point A is (3/4, 25/8).

The coordinates of point d are (0,2).

S quadrilateral adbc = s △ BOD+s △ AOD+s △ AOC =1/2+3/4+25/8 =19/4.