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a^{2}-2a-2=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.
a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)}}{2}
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.
a=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)}}{2}
-2 kvadratini chiqarish.
a=\frac{-\left(-2\right)±\sqrt{4+8}}{2}
-4 ni -2 marotabaga ko'paytirish.
a=\frac{-\left(-2\right)±\sqrt{12}}{2}
4 ni 8 ga qo'shish.
a=\frac{-\left(-2\right)±2\sqrt{3}}{2}
12 ning kvadrat ildizini chiqarish.
a=\frac{2±2\sqrt{3}}{2}
-2 ning teskarisi 2 ga teng.
a=\frac{2\sqrt{3}+2}{2}
a=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{3} ga qo'shish.
a=\sqrt{3}+1
2+2\sqrt{3} ni 2 ga bo'lish.
a=\frac{2-2\sqrt{3}}{2}
a=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{3} ni ayirish.
a=1-\sqrt{3}
2-2\sqrt{3} ni 2 ga bo'lish.
a^{2}-2a-2=\left(a-\left(\sqrt{3}+1\right)\right)\left(a-\left(1-\sqrt{3}\right)\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 1+\sqrt{3} ga va x_{2} uchun 1-\sqrt{3} ga bo‘ling.