Question

(x+4)(x+5)(x+7)(x+8)=4

Answer 1

(X^2+8x+4x+32)(x^2+5x+7x+35)=4
(X^2+12x+32)(x^2+12x+35)=4
Примем T=x^2+12x+32
T(t+3)=4
T^2+3t-4=0
D=9+16=5^2
T1=(-3-5)/2=-4
T2=(-3+5)/2=1

Для T1:
X^2+12x+32=-4
X^2+12x+36=0
D=144-4*36=0
X=-12/2=-6

Для T2:
X^2+12x+32=1
X^2+12x+31=0
D=144-4*31=20
X1=(-12-корень(20))/2
Х2=(-12+корень(20))/2