Question

(x^2-2x)^2+3(x^2-2x)-4=0 Помогите решить

Answer 1

$(x^2-2x)^2+3(x^2-2x)-4=0\\\\ x^2-2x=t\\\\ t^2+3t-4=0\\\\ t^2+4t-t-4=0\\\\ t*(t+4)-(t+4)=0\\\\ (t-1)*(t+4)=0\\\\ t-1=0\ \ or\ \ t+4=0\\\\ x^2-2x-1=0\ \ or\ \ x^2-2x+4=0\\\\ D=4+4=8\ \ \ \ \ \ \ \ \ D=4-4*4\ \textless \ 0\\\\ x_{1,2}=\frac{2\pm\sqrt{8}}{2}\\\\ x_{1,2}=\frac{2\pm2\sqrt{2}}{2}\\\\ x_{1,2}=1\pm\sqrt{2}$

Ответ: $1\pm\sqrt{2}$