# Equilibrium calculation gives negative molarity

There is a reaction: $$\ce {ClF3 <=> ClF + F2}$$ 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: $\ce {ClF3 <=> ClF + F2}$

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?