Question

Решите уравнение (x^2+6x+1)^2+3x^2+18x=1

Answer 1

$(x^2+6x+1)^2+3(x^2+6x)=1$
$x^2+6x=t$
$(t+1)^2+3t=1$
$t^2+2t+1+3t-1=0$
$t^2+5t=0$
$t(t+5)=0$
$t1=0;t2=-5$
$x^2+6x=0$
$x(x+6)=0$
$x1=0;x2=-6$
$x^2+6x=-5$
x3=-1;x4=-5
ответ:-6;-5;-1;0