Решите уравнение(х-4)^2(х+10)=15(х-4)

[latex](x-4)^{2}(x+10)=15(x-4) (x-4)^{2}(x+10)-15(x-4)=0 (x-4)((x-4)(x+10)-15)=0 (x-4)( x^{2} +6x-40-15)=0 (x-4)( x^{2}+6x-55 )=0 x_{1}=4 \sqrt{D} = \sqrt{256}=16 x_{2}=5 x_{3}=-11[/latex]

[latex](x-4)^2(x+10)=15(x-4)\\(x-4)^2(x+10)-15(x-4)=0\\(x-4)((x-4)(x+10)-15)=0\\(x-4)(x^2+6x-55)=0\\x^2+6x-55=0\ \ \ \ \ \ \ \ x-4=0\\x_{1,2}=-3^+_-8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_3=4\\x_1=-11\ x_2=5\\OTBET:x=-11,x=4,x=5[/latex]