0=2n^2+3n-340 eching

$$0 = 2 n ^ { 2 } + 3 n - 340$$

n uchun yechish

$$n = \frac{\sqrt{2729} - 3}{4} \approx 12,309957887$$

$$n=\frac{-\sqrt{2729}-3}{4}\approx -13,809957887$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2n^{2}+3n-340=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 3 ni b va -340 ni c bilan almashtiring.

$$n=\frac{-3±\sqrt{3^{2}-4\times 2\left(-340\right)}}{2\times 2}$$

3 kvadratini chiqarish.

$$n=\frac{-3±\sqrt{9-4\times 2\left(-340\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{9-8\left(-340\right)}}{2\times 2}$$

-8 ni -340 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{9+2720}}{2\times 2}$$

9 ni 2720 ga qo'shish.

$$n=\frac{-3±\sqrt{2729}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$n=\frac{-3±\sqrt{2729}}{4}$$

n=\frac{-3±\sqrt{2729}}{4} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{2729} ga qo'shish.

$$n=\frac{\sqrt{2729}-3}{4}$$

n=\frac{-3±\sqrt{2729}}{4} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{2729} ni ayirish.

$$n=\frac{-\sqrt{2729}-3}{4}$$

Tenglama yechildi.

$$n=\frac{\sqrt{2729}-3}{4} n=\frac{-\sqrt{2729}-3}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2n^{2}+3n-340=0$$

340 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$2n^{2}+3n=340$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2n^{2}+3n}{2}=\frac{340}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$n^{2}+\frac{3}{2}n=\frac{340}{2}$$

340 ni 2 ga bo'lish.

$$n^{2}+\frac{3}{2}n=170$$

\frac{3}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{4} olish uchun. Keyin, \frac{3}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$n^{2}+\frac{3}{2}n+\left(\frac{3}{4}\right)^{2}=170+\left(\frac{3}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{4} kvadratini chiqarish.

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=170+\frac{9}{16}$$

170 ni \frac{9}{16} ga qo'shish.

$$n^{2}+\frac{3}{2}n+\frac{9}{16}=\frac{2729}{16}$$

n^{2}+\frac{3}{2}n+\frac{9}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(n+\frac{3}{4}\right)^{2}=\frac{2729}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(n+\frac{3}{4}\right)^{2}}=\sqrt{\frac{2729}{16}}$$

Qisqartirish.

$$n+\frac{3}{4}=\frac{\sqrt{2729}}{4} n+\frac{3}{4}=-\frac{\sqrt{2729}}{4}$$

Tenglamaning ikkala tarafidan \frac{3}{4} ni ayirish.

$$n=\frac{\sqrt{2729}-3}{4} n=\frac{-\sqrt{2729}-3}{4}$$