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x\left(x+22\right)
x omili.
x^{2}+22x=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{-22±\sqrt{22^{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{-22±22}{2}
22^{2} ning kvadrat ildizini chiqarish.
x=\frac{0}{2}
x=\frac{-22±22}{2} tenglamasini yeching, bunda ± musbat. -22 ni 22 ga qo'shish.
x=0
0 ni 2 ga bo'lish.
x=-\frac{44}{2}
x=\frac{-22±22}{2} tenglamasini yeching, bunda ± manfiy. -22 dan 22 ni ayirish.
x=-22
-44 ni 2 ga bo'lish.
x^{2}+22x=x\left(x-\left(-22\right)\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -22 ga bo‘ling.
x^{2}+22x=x\left(x+22\right)
p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.