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F(x)=ax²+bx+c
F(5)=25a+5b+c=6 ===> c=6-5(5a-b)
Xw=(-b)/2a=3 ===>b=-6a
Yw=(-delta)/4a=(-2)
delta=b²-4ac
No i podstawiamy;]
-{(-6a)²-4a[6-5(5a-6a)]} / 4a = (-2)
O ile nie zrobilem błedu z tego wychodzi
a=1/5
b=-(6/5)
c=11
Pozdro