-m^2+2=0

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Solution for -m^2+2=0 equation:



-m^2+2=0
We add all the numbers together, and all the variables
-1m^2+2=0
a = -1; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-1)·2
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-1}=\frac{0-2\sqrt{2}}{-2} =-\frac{2\sqrt{2}}{-2} =-\frac{\sqrt{2}}{-1} $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-1}=\frac{0+2\sqrt{2}}{-2} =\frac{2\sqrt{2}}{-2} =\frac{\sqrt{2}}{-1} $

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