shear and moment diagram

First, it looks like this problem was crafted very poorly.
 

By inspection, I don't think the triangular reaction at AB will be equal to the triangular reaction at CD. The figure however suggest that they are equal because both are labeled with w_o. If you sum up forces in the vertical direction, all of the 50 kN get cancelled. What remains is the the spandrel load, the unsymmetrical placement of this load will make the triangular reactions to be unequal.
 

Another observation is that the origin for 6x² is unclear in the given. Is x measured from A or from B? Note that the load can be zero at B but does not necessarily mean to be the vertex of the parabolic curve.
 

With the errors in the given pointed out, I will attempt to answer this problem with the following premise:

• I will call the w_o at AB to be w₁ and at CD may I call it w₂.
• I am assuming that the load w has a unit of kN/m.
• I will also assume that the vertex of 6x² is at B so that at C, the magnitude is 96 kN/m

 

Here is my solution:
$\Sigma M_A = 0$

$\frac{2}{3} \cdot \frac{1}{2}(1)(w_1) + (5 + \frac{1}{3}) \cdot \frac{1}{2}(1)(w_2) = \frac{1}{3}(4)(96)\left[ 5 - \frac{1}{4}(4) \right]$

$\frac{1}{3}w_1 + \frac{8}{3}w_2 = 512$   ←   Equation (1)
 

$\Sigma M_D = 0$

$(5 + \frac{1}{3}) \cdot \frac{1}{2}(1)(w_1) + \frac{2}{3} \cdot \frac{1}{2}(1)(w_2) = \frac{1}{3}(4)(96)\left[ 1 + \frac{1}{4}(4) \right]$

$\frac{8}{3}w_1 + \frac{1}{3}w_2 = 256$   ←   Equation (2)
 

From Equations (1) and (2)
$w_1 = 73.14 \text{ kN/m}$

$w_2 = 182.86 \text{ kN/m}$