Question

5-2х2+х=0 решите уравнение

Answer 1

$\displaystyle 5-2x^2+x=0\\-2x^2+x=0\\-2x^2+x+5=0\\\\x=\frac{-1б\sqrt{1^2-4(-2)*5} }{2(-2)}=\frac{-1б\sqrt{1-4(-2)*5} }{2(-2)}=-\frac{-1б\sqrt{1+40} }{4}=-\frac{-1б\sqrt{41} }{4}=\frac{1б\sqrt{41} }{4}$