m-2m^2=-m(2-2m)

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

m-2m^2=-m(2-2m)

We move all terms to the left:

m-2m^2-(-m(2-2m))=0

We add all the numbers together, and all the variables

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


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

-m(-2m+2)

We multiply parentheses

2m^2-2m

Back to the equation:

-(2m^2-2m)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-4m^2+3m=0

a = -4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-4)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-4}=\frac{-6}{-8}
=3/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-4}=\frac{0}{-8}
=0
$