Вопрос школьника по предмету Алгебра
Помогите решить пожалуйста x²+2x-48<0 только что б понятно было)
Ответ учителя по предмету Алгебра
а>0, ветви вверх
Приравниваем к нулю
х^2+2х-48=0
а=1, b=2, c=-48
D=b^2-4ac=4+4*1*48=196=14^2>0, 2 корня
x1= (-b+конень D)/2a= (-2+14)/2= 6
x2= (-b-кореньD)/2a = (-2-14)/2=-8
Потом решаем параболой и получаем
(-8;6)
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12+(-20)-(-11)-(-6)-10=
3 5/12-(-1 1/3)+(-4 3/8)