$If(x+2a)(x+3b)=chasroots\alpha ,\beta ,thentherootsof(x+\alpha )(x+\beta )+c=0are$

$\begin{array}{l}GivenEquation(x+2a)(x+3b)=c\\ \Rightarrow x^2+(2a+3b)x+6ab-c=0\\ If\alpha ,\beta arerootsthen\\ \alpha +\beta =-(2a+3b),\alpha \beta =6ab-c\\ and(x+\alpha )(x+\beta )+c=0\\ \Rightarrow x^2+(\alpha +\beta )x+\alpha \beta +c=0\\ \Rightarrow x^2-(2a+3b)x+6ab=0\\ \Rightarrow x^2-2ax-3bx+6ab=0\\ \Rightarrow x(x-2a)-3b(x-2a)=0\\ \Rightarrow (x-2a)(x-3b)=0\\ \Rightarrow x=2a,3b\end{array}$