$\Sigma M_{R2} = 0$
$8R_1 = 4600 + \frac{1}{2}(3)(1200)(1)$
$R_1 = 800 ~ \text{N}$
$\Sigma M_{R1} = 0$
$8R_2 + 4600 = \frac{1}{2}(3)(1200)(7)$
$R_2 = 1000 ~ \text{N}$
$M_C = 4(800) - 4600 = -1400 ~ \text{N}\cdot\text{m}$
$\dfrac{y}{x - 1} = \dfrac{1200}{3}$
$y = 400(x - 1)$
Apply 3-Moment Equation to Span A-C-B
$M_A L_{AC} + 2M_C(L_{AC} + L_{CB}) + M_B L_{CB} + \left( \dfrac{6A \bar{a}}{L} \right)_{AC} + \left( \dfrac{6A \bar{b}}{L} \right)_{BC}$
$= 6EI \left( \dfrac{h_{CA}}{L_{AC}} + \dfrac{h_{CB}}{L_{BC}} \right)$
$0 + 2(-1400)(4 + 4) + 0 + \dfrac{M}{L}(3a^2 - L^2) + \sum \dfrac{Pb}{L}(L^2 - b^2)$
$= 6EI \left( \dfrac{h_C}{4} + \dfrac{h_C}{4} \right)$
$\displaystyle -22,400 + \dfrac{4600}{4}\left[ 3(2^2) - 4^2 \right] + \int_1^4 \dfrac{(y \, dx)(4 - x)}{4}\left[ 4^2 - (4 - x)^2 \right]$
$= 3EI \, h_C$
$\displaystyle -22,400 - 4600 + \dfrac{1}{4}\int_1^4 400(x - 1)(4 - x) \left[ 16 - (4 - x)^2 \right] \, dx = 3EI \, h_C$
$-22,400 - 4,600 + 5,985 = 3EI \, h_C$
$3EI \, h_C = -21,015$
$EI \, h_C = -7,005$ ← the negative sign indicates that A and B are below C
$\delta_C = \dfrac{7,005}{EI} ~ \text{upward}$ answer
