help me please

$$\small{\text{$ \begin{array}{rcl} \boxed{~~ax^2+bx+c=0 \qquad x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~} \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\ x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\ x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\ x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\ x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\ x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\ x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\ x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\ x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\ x_1 &=& \dfrac{ 2-m }{2m} \\\\ x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\ x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\ x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\ x_2 &=& \dfrac{ m }{2m} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}} \end{array} $}}$$

$${{\mathtt{mx}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{\left({\mathtt{m}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{4}}}} = {\mathtt{0}}$$

integer solutions :   m = -2       x = 1

                               m = 2        x = 0

geno has already answered this here: http://web2.0calc.com/questions/help-me-please_38 

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Sabi92, if you do not understand Geno's answer, you need to post questions about it on the original thread. :)

 x = $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$       und      x = $${\frac{{\mathtt{1}}}{{\mathtt{m}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Gruß radix !

Oh sorry i haven't seen his answer

Why didn't you see his answer?

The fact that you had an answer would have been clearly displayed in your watch list.  :/

yes but i think i confused it with other answer and reloaded the browser whithout noticed that its exactly the answer i need

 i think its because my watchlist is a bit messy with all of my posts and i should remove some of them from my watchlist

Ok, it is not a problem sabi,  

I am sure you will get yourself better organised.