Solve the initial value problem.

$\frac{d^{2}y}{d{x}^2}=3-2x,y^{\text{'}}(0)=8,y(0)=2$

A$.y=2$

B$.y=3x^2+2x^3+8x+2$

C$.y=\frac{3}{2}{x}^2+\frac{1}{3}{x}^3-8x-2$

D$.y=\frac{3}{2}{x}^2-\frac{1}{3}{x}^3+8x+2$