Question

(y-8) (2y-1)-(3y+1) (6y+2) помогите

Answer 1

(y-8) (2y-1)-(3y+1) (6y+2)= (2y^2-y-16y+8) - (18y^2+6y+6y+2)= 2y^2-17y+8-18y^2-12y-2=-16y^2-29y+6

Answer 2

$(y-8) (2y-1)-(3y+1) (6y+2)= ?$

// Раскрываем скобки

$(y-8) (2y-1)-(3y+1) (6y+2)=(2y^{2}-y-16y+8)-(18y^{2}+6y+6y+2)=2y^{2}-y-16y+8-18y^{2}-6y-6y-2=-16y^{2}-29y+6$

// Дискриминант

$-16y^{2}-29y+6=0\\a=-16\\b=-29\\c=6\\D=(-29)^2-4*(-16)*6=1225\\D=35\\y1= \frac{-29+35}{2*(-16)}=\frac{6}{-32}=\frac{3}{-16}\\y2= \frac{-29-35}{2*(-16)}=\frac{-64}{-32} =2$

Ответ: -3/16; 2