How do you solve $m^{2} + 7 = 6$ $m^2+7=6$ ?

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Answer 1 · 2017-01-28T05:36:44

$m = \pm i$ $m=+-i$

$m^{2} + 7 = 6$ $m^2+7=6$ $Subtract 6 from both sides$ $"Subtract 6 from both sides"$

$m^{2} + 7 - 6 = 6 - 6$ $m^2+7-6=6-6$ , i.e. $m^{2} + 1 = 0$ $m^2+1=0$

So, $m^{2} + 1 = 0$ $m^2+1=0$

$m^{2} = - 1$ $m^2=-1$

$m = \pm \sqrt{- 1} = \pm i$ $m=+-sqrt(-1)=+-i$

We see clearly that this has roots at $m = \pm i$ $m=+-i$ . Do you agree?

How do you solve m2+7=6m^2+7=6?