Is 'AD' the height of the trapezium? [Unless AD is the height, the end result may not be possible to show].
Area of a trapezium = (1/2) x (height) x (Sum of the two parallel sides)
==> Area here = (1/2) x AD x (AB + CD)
==> (1/2)(x-1)(3x + 2 + x + 3) = 10
==> (x - 1)(4x + 5) = 20
Expanding and simplifying, 4x^2 + x - 25 = 0