Question

(x^2+7x+13)^2-(x+3)(x+4)=1

Answer 1

$(x^2+7x+13)^2-(x+3)(x+4)=1 \ (x^2+7x+13)^2-(x^2+7x+12)=1 \ x^2+7x+12=a \ (a+1)^2-a-1=0 \ a^2+2a+1-a-1=0 \ a^2+a=0 \ a_1=0 \ a+1=0 \ a_2=-1$

$x^2+7x+12=0 \ (x+3)(x+4)=0 \ x_1=-3 \ x_2=-4 \\ x^2+7x+12=-1 \ x^2+7x+13=0 \ D=49-52<0$

Ответ: -3 и -4