Question

1)х²-4х+3=0 2)2х²+5х-3=0 3)3х²-х+2=0
ПОМОГИТЕ ПЛИИИИИЗ​

Answer 1

$\displaystyle 1)x^2-4x+3=0\\x^2-x-3x+3=0\\x(x-1)-3(x-1)=0\\(x-1)(x-3)=0\\x-1=0,x-3=0\\x_1=1,x_2=3\\\\ 2)2x^2+5x-3=0\\2x^2+6x-x-3=0\\2x(x+3)-(x+3)=0\\(x+3)(2x-1)=0\\x+3=0,2x-1=0\\\\ x_1=-3,x_2=\frac{1}{2}=0,5\\\\ 3)3x^2-x+2=0\\\\ x=\frac{-(-1)б\sqrt{(-1)^2-4*3*2} }{2*3}=\frac{1б\sqrt{1-24} }{6}=\frac{1б\sqrt{-23} }{6}\\\\ x \notin R$