Ответ: x₁=4,  x₂=-7/3.

Объяснение:

[tex]\displaystyle\\b_1=x-3\ \ \ \ \ \ b_2=x+1\ \ \ \ \ \ b_3=4x+9\\\\\frac{b_2}{b_1}=\frac{b_3}{b_2} \ \ \ \ \Rightarrow\\\\b_2^2=b_1*b_3\\\\(x+1)^2=(x-3)*(4x+9)\\\\x^2+2x+1=4x^2+9x-12x-27\\\\x^2+2x+1=4x^2-3x-27\\\\3x^2-5x-28=0\\\\3x^2-12x+7x-28=0\\\\x*(x-4)+7*(x-4)=0\\\\(x-4)*(3x+7)=0\\\\x-4=0\\\\x_1=4.\\\\3x+7=0\\\\3x=-7\ |:3\\\\x_2=-\frac{7}{3}.[/tex]