(x^2+4c)(x^2+4x-17)+60=0

Question

(x^2+4c)(x^2+4x-17)+60=0

Answer 1

(x^2 + 4x)(x^2 + 4x - 17) + 60 = 0
Замена: x^2 + 4x = t
t(t - 17) + 60 = 0
t^2 - 17t + 60 = 0
(t - 5)(t - 12) = 0
t1 = 5 ; t2 = 12

x^2 + 4x = 5 ; x^2 + 4x = 12
x^2 + 4x - 5 = 0 ; x^2 + 4x - 12 = 0
D = 4^2 - 4*1*(-5) = 36 = 6^2 ;
D = 4^2 - 4*1*(-12) = 64 = 8^2

x1 = -4 - 6 / 2 = -5
x2 = -4 + 6 /2 = 1
x3 = -4 + 8 / 2 = 2
x4 = -4 - 8 / 2 = -6