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x^{2}+3x+1=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{-3±\sqrt{3^{2}-4}}{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{-3±\sqrt{9-4}}{2}
3 kvadratini chiqarish.
x=\frac{-3±\sqrt{5}}{2}
9 ni -4 ga qo'shish.
x=\frac{\sqrt{5}-3}{2}
x=\frac{-3±\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{5} ga qo'shish.
x=\frac{-\sqrt{5}-3}{2}
x=\frac{-3±\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{5} ni ayirish.
x^{2}+3x+1=\left(x-\frac{\sqrt{5}-3}{2}\right)\left(x-\frac{-\sqrt{5}-3}{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.