(x+a)(x^2-bx)+c(x+a)=0

$\mathtt{(x+a)(x^2-bx)+c(x+a)=0;~(x+a)(x^2-bx+c)=0}\\\\\mathtt{1)~x+a=0~\to~x=-a}\\\mathtt{2)~x^2-bx+c=0;~D=b^2-4c:}\\\mathtt{1.~D\ \textgreater \ 0~\to~x_{1,2}=\frac{bб\sqrt{b^2-4c}}{2};}\\\mathtt{2.~D=0~\to~x_{1,2}=\frac{b}{2};}\\\mathtt{3.~D\ \textless \ 0~\to~HET~KOPHE`u}$

ОТВЕТ: $\mathtt{x=-a;~if~D\in[0;+\infty)~then~x=\frac{bб\sqrt{b^2-4c}}{2}~and~x=\frac{b}{2}}$