Find [m] such that the sum of solutions of the equation

$\displaystyle

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

$

is equal to 0, and the product is equal to 4.

My solution:

$\displaystyle

x1+x2=-\frac{b}{a}

$

$\displaystyle

x1*x2=\frac{c}{a}

$

$\displaystyle

x1+x2=-(2+m-m^2)

$

$\displaystyle

x1*x2=m^2

$

$\displaystyle

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

$

$\displaystyle

m^2=4

$

$\displaystyle

m=\sqrt{4}

$

$\displaystyle

m=2

$

But $\displaystyle x^2+(2+2-2^2)*x+2^2=0 $ no real solution.

How to solve this?