Roots of Cubic

Complex roots occur in conjugate pairs,so if $\color{#20A900}{1+2i}$ is a root then $\color{#20A900}{1-2i}$ is also a root.Let $a,b,c$ be the roots of the equation.By Vieta,s formulas we $\color{#D61F06}{a+b+c=-A,ab+ac+bc=B,abc=10}$ .So $\color{#624F41}{(1+2i)(1-2i)c=5c=10\rightarrow c=\frac{10}{5}=2}$ .So 3rd root is 2.So $\color{#3D99F6}{A=a+b+c=(1+2i)+(1-2i)+2=1+1+2=4\rightarrow -A=-4,B=ab+ac+bc=(1+2i)(1-2i)+(1-2i)2+(1+2i)2=5+2+2-4i+4i=9}$ .So the equation is $\color{#D61F06}{x^3-4x^2+9x-10\rightarrow B-A=9-(-4)=9+4=\boxed{13}}$