Question

Решите уравнение: х(2х-1)=3х+4

Answer 1

2(2х-1)=3х+4

4х-2=3х+4

х=6

Answer 2

$x(2x-1)=3x+4\2x^{2}-x=3x+4\2x^{2}-x-3x-4=0\2x^{2}-4x-4=0\x^{2}-2x-2=0\a=1, b=-2, c=-c\D=b^{2}-4ac=(-2)^{2}-4*1*(-2)=4+8=12\x_{1}=frac{-b-sqrt{D}}{2a}=frac{-(-2)-sqrt{12}}{2*1}=frac{2-sqrt{12}}{2}=-sqrt{12}\x_{2}=frac{-b+sqrt{D}}{2a}=frac{-(-2)+sqrt{12}}{2*1}=frac{2+sqrt{12}}{2}=sqrt{12}$