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Whakaoti mō x (complex solution)
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Tohaina

-3x^{2}+2x-4=0
Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.
x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}
Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, 2 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}
Pūrua 2.
x=\frac{-2±\sqrt{4+12\left(-4\right)}}{2\left(-3\right)}
Whakareatia -4 ki te -3.
x=\frac{-2±\sqrt{4-48}}{2\left(-3\right)}
Whakareatia 12 ki te -4.
x=\frac{-2±\sqrt{-44}}{2\left(-3\right)}
Tāpiri 4 ki te -48.
x=\frac{-2±2\sqrt{11}i}{2\left(-3\right)}
Tuhia te pūtakerua o te -44.
x=\frac{-2±2\sqrt{11}i}{-6}
Whakareatia 2 ki te -3.
x=\frac{-2+2\sqrt{11}i}{-6}
Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{11}i}{-6} ina he tāpiri te ±. Tāpiri -2 ki te 2i\sqrt{11}.
x=\frac{-\sqrt{11}i+1}{3}
Whakawehe -2+2i\sqrt{11} ki te -6.
x=\frac{-2\sqrt{11}i-2}{-6}
Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{11}i}{-6} ina he tango te ±. Tango 2i\sqrt{11} mai i -2.
x=\frac{1+\sqrt{11}i}{3}
Whakawehe -2-2i\sqrt{11} ki te -6.
x=\frac{-\sqrt{11}i+1}{3} x=\frac{1+\sqrt{11}i}{3}
Kua oti te whārite te whakatau.
-3x^{2}+2x-4=0
Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.
-3x^{2}+2x-4-\left(-4\right)=-\left(-4\right)
Me tāpiri 4 ki ngā taha e rua o te whārite.
-3x^{2}+2x=-\left(-4\right)
Mā te tango i te -4 i a ia ake anō ka toe ko te 0.
-3x^{2}+2x=4
Tango -4 mai i 0.
\frac{-3x^{2}+2x}{-3}=\frac{4}{-3}
Whakawehea ngā taha e rua ki te -3.
x^{2}+\frac{2}{-3}x=\frac{4}{-3}
Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.
x^{2}-\frac{2}{3}x=\frac{4}{-3}
Whakawehe 2 ki te -3.
x^{2}-\frac{2}{3}x=-\frac{4}{3}
Whakawehe 4 ki te -3.
x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{4}{3}+\left(-\frac{1}{3}\right)^{2}
Whakawehea te -\frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{3}. Nā, tāpiria te pūrua o te -\frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{4}{3}+\frac{1}{9}
Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{11}{9}
Tāpiri -\frac{4}{3} ki te \frac{1}{9} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.
\left(x-\frac{1}{3}\right)^{2}=-\frac{11}{9}
Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{9}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{11}{9}}
Tuhia te pūtakerua o ngā taha e rua o te whārite.
x-\frac{1}{3}=\frac{\sqrt{11}i}{3} x-\frac{1}{3}=-\frac{\sqrt{11}i}{3}
Whakarūnātia.
x=\frac{1+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i+1}{3}
Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.