Найдите корни квадратного трехчлена номер 272 полностью
1) y²-6y-7 = 0
D = b²-4ac = 64
x₁.₂ = -b±√D/2a
x1 = 7
x2 = -1
2) y²+7y+10 = 0
D = b²-4ac = 9
x1 = 1
x2 = -8
3) 2y²-y-6 = 0
D = b²-4ac = 25
x1 = 13
x2 = -12
4) 3y²+10y-8 = 0
D = b²-4ac = 196
x1 = --4
x2 = 2/3
5) 6y²+7y-3 = 0
D = b²-4ac = 121
x1 = 1/3
x2 = -1,5
6) 4y²+28y+49 = 0
D = b²-4ac = 0
x1 = -3,5
x2 = --3,5