5x^2+8x-7 eching

$$5 x ^ { 2 } + 8 x - 7$$

Omil

$$5\left(x-\frac{-\sqrt{51}-4}{5}\right)\left(x-\frac{\sqrt{51}-4}{5}\right)$$

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.

$$5x^{2}+8x-7=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 5\left(-7\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-20\left(-7\right)}}{2\times 5}$$

-20 ni -7 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+140}}{2\times 5}$$

64 ni 140 ga qo'shish.

$$x=\frac{-8±\sqrt{204}}{2\times 5}$$

204 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{51}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-8±2\sqrt{51}}{10}$$

x=\frac{-8±2\sqrt{51}}{10} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{51} ga qo'shish.

$$x=\frac{2\sqrt{51}-8}{10}$$

-8+2\sqrt{51} ni 10 ga bo'lish.

$$x=\frac{\sqrt{51}-4}{5}$$

x=\frac{-8±2\sqrt{51}}{10} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{51} ni ayirish.

$$x=\frac{-2\sqrt{51}-8}{10}$$

-8-2\sqrt{51} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{51}-4}{5}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-4+\sqrt{51}}{5} ga va x_{2} uchun \frac{-4-\sqrt{51}}{5} ga bo‘ling.

$$5x^{2}+8x-7=5\left(x-\frac{\sqrt{51}-4}{5}\right)\left(x-\frac{-\sqrt{51}-4}{5}\right)$$

Baholash

$$5x^{2}+8x-7$$