Rozwiąż równanie:
3x^{2}-19x+16=0
Δ=
Δ= •3•16
Δ= 361- 192
Δ= 169 > 0
x₁=
x₂=
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3x^{2}-19x+16=0
Δ=![b^{2}- 4ac b^{2}- 4ac](https://tex.z-dn.net/?f=b%5E%7B2%7D-+4ac)
Δ=
•3•16
Δ= 361- 192
Δ= 169 > 0
x₁=![\frac{-b-\sqrt{Δ}}{2a} \frac{-b-\sqrt{Δ}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b-%5Csqrt%7B%CE%94%7D%7D%7B2a%7D)
x₁=![\frac{19-13}{6}= 1 \frac{19-13}{6}= 1](https://tex.z-dn.net/?f=%5Cfrac%7B19-13%7D%7B6%7D%3D+1)
x₂=![\frac{-b+\sqrt{Δ}}{2a} \frac{-b+\sqrt{Δ}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%2B%5Csqrt%7B%CE%94%7D%7D%7B2a%7D)
x₂=![\frac{19+13}{6}= 5,(3) \frac{19+13}{6}= 5,(3)](https://tex.z-dn.net/?f=%5Cfrac%7B19%2B13%7D%7B6%7D%3D+5%2C%283%29)