Question

Решите уравнение
2(x-2)во2степени-x(x+1,5)=4

Answer 1

$2(x-2)^2-x(x+1,5)=4\\2(x^2-4x+4)-x^2-1,5x-4=0\\x^2-9,5x+4=0|cdot 2\\2x^2-19x+8=0\\D=361-4cdot 2cdot 8=297; ;\\ srt{D}=sqrt{297}=3sqrt{33}\\x_1=frac{19-sqrt{297}}{4}; ;; x_2=frac{19+sqrt{297}}{4}; .$