m-2m^2=-m(2-2m)-4

Solution for m-2m^2=-m(2-2m)-4 equation:

m-2m^2=-m(2-2m)-4

We move all terms to the left:

m-2m^2-(-m(2-2m)-4)=0

We add all the numbers together, and all the variables

-2m^2+m-(-m(-2m+2)-4)=0


We calculate terms in parentheses: -(-m(-2m+2)-4), so:

-m(-2m+2)-4

We multiply parentheses

2m^2-2m-4

Back to the equation:

-(2m^2-2m-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-4m^2+3m+4=0

a = -4; b = 3; c = +4;
Δ = b2-4ac
Δ = 32-4·(-4)·4
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{73}}{2*-4}=\frac{-3-\sqrt{73}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*-4}=\frac{-3+\sqrt{73}}{-8}
$