Question

2к²-12к+7=0 решите уравнение​

Answer 1

Ответ:

D=36-14=22

x1=(6-√22)/2

x2=(6+√22)/2

Answer 2

$2k^{2} -12k+7=0\\\\a=2,b=-12,c=7$

k=-(-12)±√(-12)²-4×2×7/2×2

k=12±√144-56/4

k=12±√88/4

k=12±2√22/4

k=12+2√22/4

k=12-2√22/4

k=6+√22/2

k=6-√22/2

k₁=6-√22/2

k₂=6+√22/2