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

Solution for -2m(m-3)=-3(1+m)+1 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-2m^2+6m-(-3(m+1)+1)=0


We calculate terms in parentheses: -(-3(m+1)+1), so:

-3(m+1)+1

We multiply parentheses

-3m-3+1

We add all the numbers together, and all the variables

-3m-2

Back to the equation:

-(-3m-2)


We get rid of parentheses

-2m^2+6m+3m+2=0

We add all the numbers together, and all the variables

-2m^2+9m+2=0

a = -2; b = 9; c = +2;
Δ = b2-4ac
Δ = 92-4·(-2)·2
Δ = 97
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{-(9)-\sqrt{97}}{2*-2}=\frac{-9-\sqrt{97}}{-4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{97}}{2*-2}=\frac{-9+\sqrt{97}}{-4}
$