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3x^{2}-9x+3=0
Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.
x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\times 3}}{2\times 3}
ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.
x=\frac{-\left(-9\right)±\sqrt{81-4\times 3\times 3}}{2\times 3}
-9 kvadratini chiqarish.
x=\frac{-\left(-9\right)±\sqrt{81-12\times 3}}{2\times 3}
-4 ni 3 marotabaga ko'paytirish.
x=\frac{-\left(-9\right)±\sqrt{81-36}}{2\times 3}
-12 ni 3 marotabaga ko'paytirish.
x=\frac{-\left(-9\right)±\sqrt{45}}{2\times 3}
81 ni -36 ga qo'shish.
x=\frac{-\left(-9\right)±3\sqrt{5}}{2\times 3}
45 ning kvadrat ildizini chiqarish.
x=\frac{9±3\sqrt{5}}{2\times 3}
-9 ning teskarisi 9 ga teng.
x=\frac{9±3\sqrt{5}}{6}
2 ni 3 marotabaga ko'paytirish.
x=\frac{3\sqrt{5}+9}{6}
x=\frac{9±3\sqrt{5}}{6} tenglamasini yeching, bunda ± musbat. 9 ni 3\sqrt{5} ga qo'shish.
x=\frac{\sqrt{5}+3}{2}
9+3\sqrt{5} ni 6 ga bo'lish.
x=\frac{9-3\sqrt{5}}{6}
x=\frac{9±3\sqrt{5}}{6} tenglamasini yeching, bunda ± manfiy. 9 dan 3\sqrt{5} ni ayirish.
x=\frac{3-\sqrt{5}}{2}
9-3\sqrt{5} ni 6 ga bo'lish.
3x^{2}-9x+3=3\left(x-\frac{\sqrt{5}+3}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3+\sqrt{5}}{2} ga va x_{2} uchun \frac{3-\sqrt{5}}{2} ga bo‘ling.