Now run the ICE table the other way: you know $K$ and the starting amounts, and you want the equilibrium amounts. First use Q to see which way the reaction goes, call the unknown change $x$, and solve. Often that means the quadratic formula.
Key idea
Write the E row in terms of $x$, substitute into the expression for $K$, and solve for $x$. Keep only the root that makes every concentration positive (and no larger than what you started with). Then check by putting the answers back into $K$.
Why it works
A quadratic has two roots, but only one describes a real mixture. The other one would use up more reactant than the flask holds, or give a negative concentration. The algebra cannot know that; you can.
Worked examples
Worked example: A perfect square
$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$, $K_c = 54.3$. Starting with $0.200\ \text{M}$ each of $\mathrm{H_2}$ and $\mathrm{I_2}$ and no HI, find the equilibrium concentrations.
- $Q = 0 < K$, so the reaction goes forward. E row: $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.200 - x$, $[\mathrm{HI}] = 2x$.
- $54.3 = \dfrac{(2x)^2}{(0.200 - x)^2}$. Both sides are squares, so take the square root: $7.37 = \dfrac{2x}{0.200 - x}$.
- $7.37(0.200) - 7.37x = 2x$, so $x = \dfrac{1.474}{9.37} = 0.157$.
- $[\mathrm{HI}] = 0.315\ \text{M}$; $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.043\ \text{M}$. Check: $(0.315)^2/(0.043)^2 \approx 54$.
Answer: $[\mathrm{HI}] = 0.315\ \text{M}$, $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.043\ \text{M}$.
Worked example: Using the quadratic formula
$\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, $K_c = 4.6 \times 10^{-3}$. A flask starts with $0.100\ \text{M}$ $\mathrm{N_2O_4}$. Find the equilibrium concentrations.
- E row: $[\mathrm{N_2O_4}] = 0.100 - x$, $[\mathrm{NO_2}] = 2x$. So $\dfrac{4x^2}{0.100 - x} = 4.6 \times 10^{-3}$.
- Rearrange: $4x^2 + (4.6 \times 10^{-3})x - 4.6 \times 10^{-4} = 0$.
- Quadratic formula with $a = 4$, $b = 4.6 \times 10^{-3}$, $c = -4.6 \times 10^{-4}$: $b^2 - 4ac = 7.38 \times 10^{-3}$, whose square root is $0.0859$.
- $x = \dfrac{-0.0046 \pm 0.0859}{8}$.
- The roots are $0.0102$ and $-0.0113$. The negative root would make $[\mathrm{NO_2}]$ negative, so $x = 0.0102$.
- $[\mathrm{NO_2}] = 0.0203\ \text{M}$, $[\mathrm{N_2O_4}] = 0.0898\ \text{M}$. Check: $(0.0203)^2/0.0898 = 4.6 \times 10^{-3}$.
Answer: $[\mathrm{NO_2}] = 0.0203\ \text{M}$, $[\mathrm{N_2O_4}] = 0.0898\ \text{M}$.
Common mistake
Keeping the root that makes a concentration negative, or larger than the amount you started with. Check every concentration in the E row after you pick $x$, and confirm that they reproduce $K$.
If the reaction goes in reverse ($Q > K$), write the change with the opposite signs: reactants $+x$, products $-$(coefficient)$x$.