How do you solve #3m^2+7=301#?

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Answer 1 · 2016-09-20T01:52:25

#m=+-7sqrt2#

#3m^2+7=301#

#3m^2 = 301-7 = 294#

#m^2 = 294/3 = 98#

#m = +-sqrt98#

When finding roots of intergers it is often useful for express the number as the product of its prime numbers.

#m =+-sqrt(2xx7xx7)#

Hence #m=+-7sqrt2#