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2x^{2}-7x-48=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-48\right)}}{2\times 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{-\left(-7\right)±\sqrt{49-4\times 2\left(-48\right)}}{2\times 2}
-7 kvadratini chiqarish.
x=\frac{-\left(-7\right)±\sqrt{49-8\left(-48\right)}}{2\times 2}
-4 ni 2 marotabaga ko'paytirish.
x=\frac{-\left(-7\right)±\sqrt{49+384}}{2\times 2}
-8 ni -48 marotabaga ko'paytirish.
x=\frac{-\left(-7\right)±\sqrt{433}}{2\times 2}
49 ni 384 ga qo'shish.
x=\frac{7±\sqrt{433}}{2\times 2}
-7 ning teskarisi 7 ga teng.
x=\frac{7±\sqrt{433}}{4}
2 ni 2 marotabaga ko'paytirish.
x=\frac{\sqrt{433}+7}{4}
x=\frac{7±\sqrt{433}}{4} tenglamasini yeching, bunda ± musbat. 7 ni \sqrt{433} ga qo'shish.
x=\frac{7-\sqrt{433}}{4}
x=\frac{7±\sqrt{433}}{4} tenglamasini yeching, bunda ± manfiy. 7 dan \sqrt{433} ni ayirish.
2x^{2}-7x-48=2\left(x-\frac{\sqrt{433}+7}{4}\right)\left(x-\frac{7-\sqrt{433}}{4}\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{7+\sqrt{433}}{4} ga va x_{2} uchun \frac{7-\sqrt{433}}{4} ga bo‘ling.