Question

Решите уравнение:
(x+3)(х-5)(х+2)(х-4)=60

Answer 1

$underbrace {(x+3)(x-5)}underbrace {(x+2)(x-4)}=60\\(x^2-2x-15)(x^2-2x-8)=60\\t=x^2-2x-15; ; Rightarrow ; ; x^2-2x-8=t+7\\t(t+7)=60\\t^2+7t-60=0\\t_1=-12; ,; ; t_2=5\\a); x^2-2x-15=-12; ,; ; x^2-2x-3=0\\x_1=-1,; x_2=3\\b); ; x^2-2x-15=5; ,; ; x^2-2x-20=0\\frac{D}{4}=1+20=21; ,; ; ; x_{3,4}=1pm sqrt{21}\\Otvet:; -1;; ; 3;; ; 1-sqrt{21};; ; 1+sqrt{21}; .$