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

Solution for (2m-2)(2m-2)+3(m-2)-m=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

(+4m^2-4m-4m+4)-1m+3m-6=0

We add all the numbers together, and all the variables

(+4m^2-4m-4m+4)+2m-6=0

We get rid of parentheses

4m^2-4m-4m+2m+4-6=0

We add all the numbers together, and all the variables

4m^2-6m-2=0

a = 4; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·4·(-2)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*4}=\frac{6-2\sqrt{17}}{8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*4}=\frac{6+2\sqrt{17}}{8}
$