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2x²+5x+2=0
Δ=25-4*2*2= 25-16=9
√Δ=√9=3
x1= x2=
x1= x2=2
odp. (x-)(x-2)
e).f(x)=-x²-16
-x²-16=0
-x²=16 /*(-1)
x²=-16
Δ<0 czyli układ jest ponad osią x
f).f(x)=4x²-4x+1
4x²-4x+1=0
Δ=16-4*4*1=0
x=4/8=1/2
(x-1/2)²
g). f(x)=-12x²-4x
-12x²-4x=0
x(-12x-4)=0
x1=0 -12x-4=0
x1=0 -12x=4 /:(-12)
x1=0 x2=- 1/3
12x(x+1/3)