Question

1) y - 10 y2 + 3 памагите пж​

Answer 1

$y-10y^2+3=0\\10y^2-y-3=0\\D=(-1)^2-4 \cdot 10 \cdot (-3)=1+120=121\\\sqrt{D}=11\\y_1=\dfrac{1+11}{20}=\dfrac{12}{20}=\dfrac{3}{5}\\y_2=\dfrac{1-11}{20}=\dfrac{-10}{20}=-\dfrac{1}{2}.$

Answer 2

Ответ:

$y-10y^2+3=-(10y^2-y-3)=-10\Big(y+\dfrac{1}{2}\Big)\Big(y-\dfrac{3}{5}\Big)=(2y+1)(3-5y)\\\\\\\star \ \ 10y^2-y-3=0\ \ ,\ \ D=1+120=121\ \ ,\\\\x_1=\dfrac{1-11}{20}=\dfrac{-10}{20}=-\dfrac{1}{2}\ \ \ ,\ \ \ x_2=\dfrac{1+11}{20}=\dfrac{12}{20}=\dfrac{3}{5}\ \ \star$