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5x^{2}-2x-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.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\left(-2\right)}}{2\times 5}
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(-2\right)±\sqrt{4-4\times 5\left(-2\right)}}{2\times 5}
-2 kvadratini chiqarish.
x=\frac{-\left(-2\right)±\sqrt{4-20\left(-2\right)}}{2\times 5}
-4 ni 5 marotabaga ko'paytirish.
x=\frac{-\left(-2\right)±\sqrt{4+40}}{2\times 5}
-20 ni -2 marotabaga ko'paytirish.
x=\frac{-\left(-2\right)±\sqrt{44}}{2\times 5}
4 ni 40 ga qo'shish.
x=\frac{-\left(-2\right)±2\sqrt{11}}{2\times 5}
44 ning kvadrat ildizini chiqarish.
x=\frac{2±2\sqrt{11}}{2\times 5}
-2 ning teskarisi 2 ga teng.
x=\frac{2±2\sqrt{11}}{10}
2 ni 5 marotabaga ko'paytirish.
x=\frac{2\sqrt{11}+2}{10}
x=\frac{2±2\sqrt{11}}{10} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{11} ga qo'shish.
x=\frac{\sqrt{11}+1}{5}
2+2\sqrt{11} ni 10 ga bo'lish.
x=\frac{2-2\sqrt{11}}{10}
x=\frac{2±2\sqrt{11}}{10} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{11} ni ayirish.
x=\frac{1-\sqrt{11}}{5}
2-2\sqrt{11} ni 10 ga bo'lish.
5x^{2}-2x-2=5\left(x-\frac{\sqrt{11}+1}{5}\right)\left(x-\frac{1-\sqrt{11}}{5}\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{1+\sqrt{11}}{5} ga va x_{2} uchun \frac{1-\sqrt{11}}{5} ga bo‘ling.