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x^{2}+4x+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{-4±\sqrt{4^{2}-4\times 2}}{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.
x=\frac{-4±\sqrt{16-4\times 2}}{2}
4 kvadratini chiqarish.
x=\frac{-4±\sqrt{16-8}}{2}
-4 ni 2 marotabaga ko'paytirish.
x=\frac{-4±\sqrt{8}}{2}
16 ni -8 ga qo'shish.
x=\frac{-4±2\sqrt{2}}{2}
8 ning kvadrat ildizini chiqarish.
x=\frac{2\sqrt{2}-4}{2}
x=\frac{-4±2\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{2} ga qo'shish.
x=\sqrt{2}-2
-4+2\sqrt{2} ni 2 ga bo'lish.
x=\frac{-2\sqrt{2}-4}{2}
x=\frac{-4±2\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{2} ni ayirish.
x=-\sqrt{2}-2
-4-2\sqrt{2} ni 2 ga bo'lish.
x^{2}+4x+2=\left(x-\left(\sqrt{2}-2\right)\right)\left(x-\left(-\sqrt{2}-2\right)\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -2+\sqrt{2} ga va x_{2} uchun -2-\sqrt{2} ga bo‘ling.