-2a^2-2a+6 eching

$$-2 { a }^{ 2 } -2a+6$$

Omil

$$-2\left(a-\frac{-\sqrt{13}-1}{2}\right)\left(a-\frac{\sqrt{13}-1}{2}\right)$$

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.

$$-2a^{2}-2a+6=0$$

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.

$$a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

-2 kvadratini chiqarish.

$$a=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$a=\frac{-\left(-2\right)±\sqrt{4+8\times 6}}{2\left(-2\right)}$$

8 ni 6 marotabaga ko'paytirish.

$$a=\frac{-\left(-2\right)±\sqrt{4+48}}{2\left(-2\right)}$$

4 ni 48 ga qo'shish.

$$a=\frac{-\left(-2\right)±\sqrt{52}}{2\left(-2\right)}$$

52 ning kvadrat ildizini chiqarish.

$$a=\frac{-\left(-2\right)±2\sqrt{13}}{2\left(-2\right)}$$

-2 ning teskarisi 2 ga teng.

$$a=\frac{2±2\sqrt{13}}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$a=\frac{2±2\sqrt{13}}{-4}$$

a=\frac{2±2\sqrt{13}}{-4} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{13} ga qo'shish.

$$a=\frac{2\sqrt{13}+2}{-4}$$

2+2\sqrt{13} ni -4 ga bo'lish.

$$a=\frac{-\sqrt{13}-1}{2}$$

a=\frac{2±2\sqrt{13}}{-4} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{13} ni ayirish.

$$a=\frac{2-2\sqrt{13}}{-4}$$

2-2\sqrt{13} ni -4 ga bo'lish.

$$a=\frac{\sqrt{13}-1}{2}$$

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{13}}{2} ga va x_{2} uchun \frac{-1+\sqrt{13}}{2} ga bo‘ling.

$$-2a^{2}-2a+6=-2\left(a-\frac{-\sqrt{13}-1}{2}\right)\left(a-\frac{\sqrt{13}-1}{2}\right)$$

Baholash

$$6-2a-2a^{2}$$