The ka of a monoprotic weak acid is 4.97 × 10-3. what is the percent ionization of a 0.193 m solution of this acid?

Given:

Ka of monoprotic weak acid = 4.97 x 10^-3
Concentration = 0.193 m 

H2A > 2H2+ + A- 
0.193         0       0
-x               x        x

0.193 -x      x        x

where x is the percent ionization

4.97 x 10^-3 = x^2 * x / 0.193 - x

solve for x

x = 0.082

What? i don’t understand