How do you solve $3 m^{2} + 7 = 301$ $3m^2+7=301$ ?

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Answer 1 · 2016-09-20T01:52:25

$m = \pm 7 \sqrt{2}$ $m=+-7sqrt2$

$3 m^{2} + 7 = 301$ $3m^2+7=301$

$3 m^{2} = 301 - 7 = 294$ $3m^2 = 301-7 = 294$

$m^{2} = \frac{294}{3} = 98$ $m^2 = 294/3 = 98$

$m = \pm \sqrt{98}$ $m = +-sqrt98$

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

$m = \pm \sqrt{2 \times 7 \times 7}$ $m =+-sqrt(2xx7xx7)$

Hence $m = \pm 7 \sqrt{2}$ $m=+-7sqrt2$

How do you solve 3m^2+7=3013m2+7=3013m^2+7=301?