6-6x^2-3x= eching

$$6 - 6 x ^ { 2 } - 3 x =$$

Omil

$$-6\left(x-\frac{-\sqrt{17}-1}{4}\right)\left(x-\frac{\sqrt{17}-1}{4}\right)$$

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.

$$-6x^{2}-3x+6=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

-4 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 6}}{2\left(-6\right)}$$

24 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+144}}{2\left(-6\right)}$$

9 ni 144 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{153}}{2\left(-6\right)}$$

153 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±3\sqrt{17}}{2\left(-6\right)}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±3\sqrt{17}}{2\left(-6\right)}$$

2 ni -6 marotabaga ko'paytirish.

$$x=\frac{3±3\sqrt{17}}{-12}$$

x=\frac{3±3\sqrt{17}}{-12} tenglamasini yeching, bunda ± musbat. 3 ni 3\sqrt{17} ga qo'shish.

$$x=\frac{3\sqrt{17}+3}{-12}$$

3+3\sqrt{17} ni -12 ga bo'lish.

$$x=\frac{-\sqrt{17}-1}{4}$$

x=\frac{3±3\sqrt{17}}{-12} tenglamasini yeching, bunda ± manfiy. 3 dan 3\sqrt{17} ni ayirish.

$$x=\frac{3-3\sqrt{17}}{-12}$$

3-3\sqrt{17} ni -12 ga bo'lish.

$$x=\frac{\sqrt{17}-1}{4}$$

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{17}}{4} ga va x_{2} uchun \frac{-1+\sqrt{17}}{4} ga bo‘ling.

$$-6x^{2}-3x+6=-6\left(x-\frac{-\sqrt{17}-1}{4}\right)\left(x-\frac{\sqrt{17}-1}{4}\right)$$

Baholash

$$6-3x-6x^{2}$$