4n^2-n-812 eching

$$4 n ^ { 2 } - n - 812$$

Omil

$$4\left(n-\frac{1-\sqrt{12993}}{8}\right)\left(n-\frac{\sqrt{12993}+1}{8}\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$4n^{2}-n-812=0$$

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.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 4\left(-812\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$n=\frac{-\left(-1\right)±\sqrt{1-16\left(-812\right)}}{2\times 4}$$

-16 ni -812 marotabaga ko'paytirish.

$$n=\frac{-\left(-1\right)±\sqrt{1+12992}}{2\times 4}$$

1 ni 12992 ga qo'shish.

$$n=\frac{-\left(-1\right)±\sqrt{12993}}{2\times 4}$$

-1 ning teskarisi 1 ga teng.

$$n=\frac{1±\sqrt{12993}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$n=\frac{1±\sqrt{12993}}{8}$$

n=\frac{1±\sqrt{12993}}{8} tenglamasini yeching, bunda ± musbat. 1 ni \sqrt{12993} ga qo'shish.

$$n=\frac{\sqrt{12993}+1}{8}$$

n=\frac{1±\sqrt{12993}}{8} tenglamasini yeching, bunda ± manfiy. 1 dan \sqrt{12993} ni ayirish.

$$n=\frac{1-\sqrt{12993}}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{1+\sqrt{12993}}{8} ga va x_{2} uchun \frac{1-\sqrt{12993}}{8} ga bo‘ling.

$$4n^{2}-n-812=4\left(n-\frac{\sqrt{12993}+1}{8}\right)\left(n-\frac{1-\sqrt{12993}}{8}\right)$$

Baholash

$$4n^{2}-n-812$$