2x^2+7x-75 eching

$$2 x ^ { 2 } + 7 x - 75$$

Omil

$$2\left(x-\frac{-\sqrt{649}-7}{4}\right)\left(x-\frac{\sqrt{649}-7}{4}\right)$$

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.

$$2x^{2}+7x-75=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\left(-75\right)}}{2\times 2}$$

7 kvadratini chiqarish.

$$x=\frac{-7±\sqrt{49-4\times 2\left(-75\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-8\left(-75\right)}}{2\times 2}$$

-8 ni -75 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49+600}}{2\times 2}$$

49 ni 600 ga qo'shish.

$$x=\frac{-7±\sqrt{649}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{649}}{4}$$

x=\frac{-7±\sqrt{649}}{4} tenglamasini yeching, bunda ± musbat. -7 ni \sqrt{649} ga qo'shish.

$$x=\frac{\sqrt{649}-7}{4}$$

x=\frac{-7±\sqrt{649}}{4} tenglamasini yeching, bunda ± manfiy. -7 dan \sqrt{649} ni ayirish.

$$x=\frac{-\sqrt{649}-7}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-7+\sqrt{649}}{4} ga va x_{2} uchun \frac{-7-\sqrt{649}}{4} ga bo‘ling.

$$2x^{2}+7x-75=2\left(x-\frac{\sqrt{649}-7}{4}\right)\left(x-\frac{-\sqrt{649}-7}{4}\right)$$

Baholash

$$2x^{2}+7x-75$$