Equilibrium calculation gives negative molarity There is a reaction: $$\ce {ClF3 &lt;=&gt; ClF + F2}$$ with $\ce {K_\mathrm{c} = 2.48 x 10^{-3}}$ Initially, $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, and $\ce {[F2] = 0.02 M} $. $\ce {F2}$ is added until the concentration of $\ce {F2} $ is $\ce {0.1 M} $.
  What is the concentration of these substances after equilibrium?


My attempt:
$\ce {ClF3 &lt;=&gt; ClF + F2} $

Initial concentration: $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, $\ce {[F2] = 0.1 M} $
Concentration that react: $\ce {[ClF3] = +x M} $, $\ce {[ClF] = -x M} $, $\ce {[F2] = -x M}$
Concentration after equilibrium: $\ce {[ClF3] = 0.25+x M} $, $\ce {[ClF] = 0.031-x M} $, $\ce {[F2] = 0.1-x M} $
So, $\ce {Kc = [ClF].[F2].[ClF3]^{-1}} $  

I applied the concentration after equilibrium in $\ce {Kc} $, so I got:
$\ce {2.48.10^{-3} = [0.1-x].[0.031-x].[0.25+x]^{-1}} $

After calculate the equation above, I got $\ce {x = 0.1111 M}$. 

Then my question is: if $\ce {x = 0.1111 M} $, so $\ce {[ClF] = -0.0801 M} $ and $\ce {[F2] = -0.0111 M} $. The concentration is negative. Is it possible? Are my steps incorrect or the answer is correct?
The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?