5t^2+10t eching

$$5 t ^ { 2 } + 10 t$$

Omil

$$5t\left(t+2\right)$$

Yechim qadamlari

5 omili.

$$5\left(t^{2}+2t\right)$$

Hisoblang: t^{2}+2t. t omili.

$$t\left(t+2\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5t\left(t+2\right)$$

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.

$$5t^{2}+10t=0$$

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.

$$t=\frac{-10±\sqrt{10^{2}}}{2\times 5}$$

10^{2} ning kvadrat ildizini chiqarish.

$$t=\frac{-10±10}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$t=\frac{-10±10}{10}$$

t=\frac{-10±10}{10} tenglamasini yeching, bunda ± musbat. -10 ni 10 ga qo'shish.

$$t=\frac{0}{10}$$

0 ni 10 ga bo'lish.

$$t=0$$

t=\frac{-10±10}{10} tenglamasini yeching, bunda ± manfiy. -10 dan 10 ni ayirish.

$$t=-\frac{20}{10}$$

-20 ni 10 ga bo'lish.

$$t=-2$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -2 ga bo‘ling.

$$5t^{2}+10t=5t\left(t-\left(-2\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$5t^{2}+10t=5t\left(t+2\right)$$

Baholash

$$5t\left(t+2\right)$$