Question

5у^2+2у-3 кто нибудь поможет​

Answer 1

Ответ:   -1.   0.6.

Объяснение:

5у^2+2у-3=0;

a=5;  b=2;  c=-3;

D=b²-4ac=2²-4*5(-3)=4+60=64>0 - 2 корня.

y1.2=(-b±√D)/2a=(-2±√64)/2*5=(-2±8)/10;

y1=(-2+8)/10=6/10=3/5=0.6;

y2=(-2-8)/10=-10/10=-1.

Answer 2

Ответ:  $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\ddddddddddddddddddd \huge \boxed{y_1=-1 \ ; \ y_2=0,6}$

Объяснение:

$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\ddddddddddddddddddd \displaystyle\ \Large \boldsymbol{} 1-st \ method\\\\5y^2+2y-3=0 \\\\ D=4+4\cdot5\cdot3=64 \\\\\\y_1= \frac{-2-8}{10} =-1 \\\\\\y_2=\frac{-2+8}{10} =0,6 \\\\---------------\\\\2-nd \ method \\\\5y^2+2y-3=0$                                                                                                                                  $\displaystyle\ \Large \boldsymbol{} \underbrace{3y^2+2y^2}_{5y^2}+2y-3=0 \\\\\\3y^2-3+2y^2+2y=0 \\\\ 3(y^2-1)+2(y^2+y)=0 \\\\3\underline{(y+1)}(y-1)+2y\underline{(y+1)}=0 \\\\\\ (y+1)(3y-3+2y)=0 \\\\ (y+1)(5y-3) \Longrightarrow y_1=-1 \ ; \ \ y_2 =0,6$