x²-4x+11=8 proszę o rozwiązanie
x²-4x+11=8 proszę o rozwiązaniex²-4x+3=0
Δ=16-4*1*3
Δ=4
√Δ=2
x₁=(4-2)/2=1
x₂=(4+2)/2=3
x²-4x+11=8
x²-4x+3=0
Δ=b²-4ac
Δ=16-12=4
x₁=(-b+√Δ)/2a=(4+2)/2=3
x₂=(-b-√Δ)/2a=(4-2)/2=1
wyliczam deltę
Δ= b²-4ac
Δ= (-4)²- 4*1*3
Δ= 16-12
Δ= 4
√Δ= 2
x₁= -b+ √Δ /2a
x₁= 4+2 /2a
x₁= 6/2
x₁= 3
x₂= -b- √Δ /2a
x₁= 4-2 /2a
x₁= 2/2
x₁= 1
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x²-4x+11=8 proszę o rozwiązanie
x²-4x+3=0
Δ=16-4*1*3
Δ=4
√Δ=2
x₁=(4-2)/2=1
x₂=(4+2)/2=3
x²-4x+11=8
x²-4x+3=0
Δ=b²-4ac
Δ=16-12=4
x₁=(-b+√Δ)/2a=(4+2)/2=3
x₂=(-b-√Δ)/2a=(4-2)/2=1
x²-4x+11=8
x²-4x+3=0
wyliczam deltę
Δ= b²-4ac
Δ= (-4)²- 4*1*3
Δ= 16-12
Δ= 4
√Δ= 2
x₁= -b+ √Δ /2a
x₁= 4+2 /2a
x₁= 6/2
x₁= 3
x₂= -b- √Δ /2a
x₁= 4-2 /2a
x₁= 2/2
x₁= 1