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x^{2}\times 10+36=4590-12x
Tenglamaning ikkala tarafini 6 ga ko'paytirish.
x^{2}\times 10+36-4590=-12x
Ikkala tarafdan 4590 ni ayirish.
x^{2}\times 10-4554=-12x
-4554 olish uchun 36 dan 4590 ni ayirish.
x^{2}\times 10-4554+12x=0
12x ni ikki tarafga qo’shing.
10x^{2}+12x-4554=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{-12±\sqrt{12^{2}-4\times 10\left(-4554\right)}}{2\times 10}
Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 10 ni a, 12 ni b va -4554 ni c bilan almashtiring.
x=\frac{-12±\sqrt{144-4\times 10\left(-4554\right)}}{2\times 10}
12 kvadratini chiqarish.
x=\frac{-12±\sqrt{144-40\left(-4554\right)}}{2\times 10}
-4 ni 10 marotabaga ko'paytirish.
x=\frac{-12±\sqrt{144+182160}}{2\times 10}
-40 ni -4554 marotabaga ko'paytirish.
x=\frac{-12±\sqrt{182304}}{2\times 10}
144 ni 182160 ga qo'shish.
x=\frac{-12±12\sqrt{1266}}{2\times 10}
182304 ning kvadrat ildizini chiqarish.
x=\frac{-12±12\sqrt{1266}}{20}
2 ni 10 marotabaga ko'paytirish.
x=\frac{12\sqrt{1266}-12}{20}
x=\frac{-12±12\sqrt{1266}}{20} tenglamasini yeching, bunda ± musbat. -12 ni 12\sqrt{1266} ga qo'shish.
x=\frac{3\sqrt{1266}-3}{5}
-12+12\sqrt{1266} ni 20 ga bo'lish.
x=\frac{-12\sqrt{1266}-12}{20}
x=\frac{-12±12\sqrt{1266}}{20} tenglamasini yeching, bunda ± manfiy. -12 dan 12\sqrt{1266} ni ayirish.
x=\frac{-3\sqrt{1266}-3}{5}
-12-12\sqrt{1266} ni 20 ga bo'lish.
x=\frac{3\sqrt{1266}-3}{5} x=\frac{-3\sqrt{1266}-3}{5}
Tenglama yechildi.
x^{2}\times 10+36=4590-12x
Tenglamaning ikkala tarafini 6 ga ko'paytirish.
x^{2}\times 10+36+12x=4590
12x ni ikki tarafga qo’shing.
x^{2}\times 10+12x=4590-36
Ikkala tarafdan 36 ni ayirish.
x^{2}\times 10+12x=4554
4554 olish uchun 4590 dan 36 ni ayirish.
10x^{2}+12x=4554
Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.
\frac{10x^{2}+12x}{10}=\frac{4554}{10}
Ikki tarafini 10 ga bo‘ling.
x^{2}+\frac{12}{10}x=\frac{4554}{10}
10 ga bo'lish 10 ga ko'paytirishni bekor qiladi.
x^{2}+\frac{6}{5}x=\frac{4554}{10}
\frac{12}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.
x^{2}+\frac{6}{5}x=\frac{2277}{5}
\frac{4554}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.
x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{2277}{5}+\left(\frac{3}{5}\right)^{2}
\frac{6}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{5} olish uchun. Keyin, \frac{3}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{2277}{5}+\frac{9}{25}
Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{5} kvadratini chiqarish.
x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{11394}{25}
Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2277}{5} ni \frac{9}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.
\left(x+\frac{3}{5}\right)^{2}=\frac{11394}{25}
x^{2}+\frac{6}{5}x+\frac{9}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.
\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{11394}{25}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x+\frac{3}{5}=\frac{3\sqrt{1266}}{5} x+\frac{3}{5}=-\frac{3\sqrt{1266}}{5}
Qisqartirish.
x=\frac{3\sqrt{1266}-3}{5} x=\frac{-3\sqrt{1266}-3}{5}
Tenglamaning ikkala tarafidan \frac{3}{5} ni ayirish.