m(2)=3-m(2)

Solution for m(2)=3-m(2) equation:

m(2)=3-m(2)

We move all terms to the left:

m(2)-(3-m(2))=0

We add all the numbers together, and all the variables

-(-1m^2+3)+m2=0

We add all the numbers together, and all the variables

m^2-(-1m^2+3)=0

We get rid of parentheses

m^2+1m^2-3=0

We add all the numbers together, and all the variables

2m^2-3=0

a = 2; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·2·(-3)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*2}=\frac{0-2\sqrt{6}}{4}
=-\frac{2\sqrt{6}}{4}
=-\frac{\sqrt{6}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*2}=\frac{0+2\sqrt{6}}{4}
=\frac{2\sqrt{6}}{4}
=\frac{\sqrt{6}}{2}
$