Finding out square root of any number?

Hi Sourabh,

Please try this :

import decimal
num = 10708709524244500001847324521347505240227690877120297398105989627045345408925857979704697480024468946600583137559854588917716112950659408968489340748064765649966448212581828337282245796573264040262525597849040830556236566097590451594821584127105495178885540154047154009432149785581739577662990157086720107087095242445000018473245213475052402276908771202973981059896270453454089258579797046974800244689466005831375598545889177161129506594089684893407480647656499664482125818283372822457965732640402625255978490408305562365660975904515948215841271054951788855401540471540094321497855817395776629901570867201070870952424450000184732452134750524022769087712029739810598962704534540892585797970469748002446894660058313755985458891771611295065940896848934074806476564996644821258182833728224579657326404026252559784904083055623656609759045159482158412710549517888554015404715400943214978558173957766299015708672010708709524244500001847324521347505240227690877120297398105989627045345408925857979704697480024468946600583137559854588917716112950659408968489340748064765649966448212581828337282245796573264040262525597849040830556236566097590451594821584127105495178885540154047154009432149785581739577662990157086720
d = decimal.Decimal(num)
d.sqrt

Reference :

https://stackoverflow.com/questions/20689683/overflowerror-long-int-too-large-to-convert-to-float

2 Replies

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Rajeesh Menoth9y ago
Hi Sourabh,

Please try this :
import decimal

num = 10708709524244500001847324521347505240227690877120297398105989627045345408925857979704697480024468946600583137559854588917716112950659408968489340748064765649966448212581828337282245796573264040262525597849040830556236566097590451594821584127105495178885540154047154009432149785581739577662990157086720107087095242445000018473245213475052402276908771202973981059896270453454089258579797046974800244689466005831375598545889177161129506594089684893407480647656499664482125818283372822457965732640402625255978490408305562365660975904515948215841271054951788855401540471540094321497855817395776629901570867201070870952424450000184732452134750524022769087712029739810598962704534540892585797970469748002446894660058313755985458891771611295065940896848934074806476564996644821258182833728224579657326404026252559784904083055623656609759045159482158412710549517888554015404715400943214978558173957766299015708672010708709524244500001847324521347505240227690877120297398105989627045345408925857979704697480024468946600583137559854588917716112950659408968489340748064765649966448212581828337282245796573264040262525597849040830556236566097590451594821584127105495178885540154047154009432149785581739577662990157086720
d = decimal.Decimal(num)
d.sqrt
Reference :
https://stackoverflow.com/questions/20689683/overflowerror-long-int-too-large-to-convert-to-float
+1
Umair Hassan9y ago
Use Math.Sqrt(Double Value) to find squre root of any number

Useful Resource:
https://msdn.microsoft.com/en-us/library/system.math.sqrt(v=vs.110).aspx
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