1+2m=2m(m+2)

Solution for 1+2m=2m(m+2) equation:

1+2m=2m(m+2)

We move all terms to the left:

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


We calculate terms in parentheses: -(2m(m+2)), so:

2m(m+2)

We multiply parentheses

2m^2+4m

Back to the equation:

-(2m^2+4m)


We get rid of parentheses

-2m^2+2m-4m+1=0

We add all the numbers together, and all the variables

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

a = -2; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-2)·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2*-2}=\frac{2-2\sqrt{3}}{-4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2*-2}=\frac{2+2\sqrt{3}}{-4}
$