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2x^{2}+5x-4=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{-5±\sqrt{5^{2}-4\times 2\left(-4\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{-5±\sqrt{25-4\times 2\left(-4\right)}}{2\times 2}
5 kvadratini chiqarish.
x=\frac{-5±\sqrt{25-8\left(-4\right)}}{2\times 2}
-4 ni 2 marotabaga ko'paytirish.
x=\frac{-5±\sqrt{25+32}}{2\times 2}
-8 ni -4 marotabaga ko'paytirish.
x=\frac{-5±\sqrt{57}}{2\times 2}
25 ni 32 ga qo'shish.
x=\frac{-5±\sqrt{57}}{4}
2 ni 2 marotabaga ko'paytirish.
x=\frac{\sqrt{57}-5}{4}
x=\frac{-5±\sqrt{57}}{4} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{57} ga qo'shish.
x=\frac{-\sqrt{57}-5}{4}
x=\frac{-5±\sqrt{57}}{4} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{57} ni ayirish.
2x^{2}+5x-4=2\left(x-\frac{\sqrt{57}-5}{4}\right)\left(x-\frac{-\sqrt{57}-5}{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{-5+\sqrt{57}}{4} ga va x_{2} uchun \frac{-5-\sqrt{57}}{4} ga bo‘ling.