Question

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

Answer 1

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

ОТВЕТ: $mathtt{x=-a;~if~Din[0;+infty)~then~x=frac{bбsqrt{b^2-4c}}{2}~and~x=frac{b}{2}}$