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

x₁.₂ = -b±√D/2a

x1 = 1

x2 = -8

3)  2y²-y-6 = 0

D = b²-4ac = 25

x₁.₂ = -b±√D/2a

x1 = 13

x2 = -12

4)  3y²+10y-8 = 0

D = b²-4ac = 196

x₁.₂ = -b±√D/2a

x1 = --4

x2 = 2/3

5)  6y²+7y-3 = 0

D = b²-4ac = 121

x₁.₂ = -b±√D/2a

x1 = 1/3

x2 = -1,5

6)  4y²+28y+49 = 0

D = b²-4ac = 0

x₁.₂ = -b±√D/2a

x1 = -3,5

x2 = --3,5