Question

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

Answer 1

$(x^2-2x)^2+3(x^2-2x)-4=0\\ x^2-2x=t\\ t^2+3t-4=0\\ t^2+4t-t-4=0\\ t*(t+4)-(t+4)=0\\ (t-1)*(t+4)=0\\ t-1=0 or t+4=0\\ x^2-2x-1=0 or x^2-2x+4=0\\ D=4+4=8 D=4-4*4 textless 0\\ x_{1,2}=frac{2pmsqrt{8}}{2}\\ x_{1,2}=frac{2pm2sqrt{2}}{2}\\ x_{1,2}=1pmsqrt{2}$

Ответ: $1pmsqrt{2}$