3b^2+15b+2 eching

$$3 b ^ { 2 } + 15 b + 2$$

Omil

$$3\left(b-\left(-\frac{\sqrt{201}}{6}-\frac{5}{2}\right)\right)\left(b-\left(\frac{\sqrt{201}}{6}-\frac{5}{2}\right)\right)$$

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.

$$3b^{2}+15b+2=0$$

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.

$$b=\frac{-15±\sqrt{15^{2}-4\times 3\times 2}}{2\times 3}$$

15 kvadratini chiqarish.

$$b=\frac{-15±\sqrt{225-4\times 3\times 2}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$b=\frac{-15±\sqrt{225-12\times 2}}{2\times 3}$$

-12 ni 2 marotabaga ko'paytirish.

$$b=\frac{-15±\sqrt{225-24}}{2\times 3}$$

225 ni -24 ga qo'shish.

$$b=\frac{-15±\sqrt{201}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$b=\frac{-15±\sqrt{201}}{6}$$

b=\frac{-15±\sqrt{201}}{6} tenglamasini yeching, bunda ± musbat. -15 ni \sqrt{201} ga qo'shish.

$$b=\frac{\sqrt{201}-15}{6}$$

-15+\sqrt{201} ni 6 ga bo'lish.

$$b=\frac{\sqrt{201}}{6}-\frac{5}{2}$$

b=\frac{-15±\sqrt{201}}{6} tenglamasini yeching, bunda ± manfiy. -15 dan \sqrt{201} ni ayirish.

$$b=\frac{-\sqrt{201}-15}{6}$$

-15-\sqrt{201} ni 6 ga bo'lish.

$$b=-\frac{\sqrt{201}}{6}-\frac{5}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{5}{2}+\frac{\sqrt{201}}{6} ga va x_{2} uchun -\frac{5}{2}-\frac{\sqrt{201}}{6} ga bo‘ling.

$$3b^{2}+15b+2=3\left(b-\left(\frac{\sqrt{201}}{6}-\frac{5}{2}\right)\right)\left(b-\left(-\frac{\sqrt{201}}{6}-\frac{5}{2}\right)\right)$$

Baholash

$$3b^{2}+15b+2$$