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3x^{2}+6x=8
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.
3x^{2}+6x-8=8-8
Tenglamaning ikkala tarafidan 8 ni ayirish.
3x^{2}+6x-8=0
O‘zidan 8 ayirilsa 0 qoladi.
x=\frac{-6±\sqrt{6^{2}-4\times 3\left(-8\right)}}{2\times 3}
Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 6 ni b va -8 ni c bilan almashtiring.
x=\frac{-6±\sqrt{36-4\times 3\left(-8\right)}}{2\times 3}
6 kvadratini chiqarish.
x=\frac{-6±\sqrt{36-12\left(-8\right)}}{2\times 3}
-4 ni 3 marotabaga ko'paytirish.
x=\frac{-6±\sqrt{36+96}}{2\times 3}
-12 ni -8 marotabaga ko'paytirish.
x=\frac{-6±\sqrt{132}}{2\times 3}
36 ni 96 ga qo'shish.
x=\frac{-6±2\sqrt{33}}{2\times 3}
132 ning kvadrat ildizini chiqarish.
x=\frac{-6±2\sqrt{33}}{6}
2 ni 3 marotabaga ko'paytirish.
x=\frac{2\sqrt{33}-6}{6}
x=\frac{-6±2\sqrt{33}}{6} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{33} ga qo'shish.
x=\frac{\sqrt{33}}{3}-1
-6+2\sqrt{33} ni 6 ga bo'lish.
x=\frac{-2\sqrt{33}-6}{6}
x=\frac{-6±2\sqrt{33}}{6} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{33} ni ayirish.
x=-\frac{\sqrt{33}}{3}-1
-6-2\sqrt{33} ni 6 ga bo'lish.
x=\frac{\sqrt{33}}{3}-1 x=-\frac{\sqrt{33}}{3}-1
Tenglama yechildi.
3x^{2}+6x=8
Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.
\frac{3x^{2}+6x}{3}=\frac{8}{3}
Ikki tarafini 3 ga bo‘ling.
x^{2}+\frac{6}{3}x=\frac{8}{3}
3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.
x^{2}+2x=\frac{8}{3}
6 ni 3 ga bo'lish.
x^{2}+2x+1^{2}=\frac{8}{3}+1^{2}
2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}+2x+1=\frac{8}{3}+1
1 kvadratini chiqarish.
x^{2}+2x+1=\frac{11}{3}
\frac{8}{3} ni 1 ga qo'shish.
\left(x+1\right)^{2}=\frac{11}{3}
x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.
\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{11}{3}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x+1=\frac{\sqrt{33}}{3} x+1=-\frac{\sqrt{33}}{3}
Qisqartirish.
x=\frac{\sqrt{33}}{3}-1 x=-\frac{\sqrt{33}}{3}-1
Tenglamaning ikkala tarafidan 1 ni ayirish.