For more ideas, see the JavaScript EquationSolving Calculator
Full solutions are not given here. Instead, I just show how the equation gets put into the TI85 and solved.
Page 554, Sample Exercise 15.10, Chemistry: The Central Science, 7/e
For the Haber process, N_{2}(g) + 3 H_{2}(g) <==> 2 NH_{3}(g) , K_{p} = 1.45E5 at 500 ^{o}C. In an equilibrium mixture of the three gases at 500 ^{o}C the partial pressure of H _{2} is 0.928 atm and that of N_{2} is 0.432 atm. What is the partial pressure of NH_{3} in this equilibrium mixture? eqn:Kp=x^2/(0.432*0.928^3) Kp=1.4E5 Kp=1.4E5 x= Press [SOLVE]==> .x=.0022374216639981 bound={1E99,1E99} bound={1E99,1E99}Answer: The equilibrium pressure of NH_{3} is 0.00224 atm.

Page 555, Sample Exercise 15.11, Chemistry: The Central Science, 7/e
A 1.000L flask is filled with 1.000 mol of H_{2} and 2.000 mol of I_{2} at 448 ^{o}C. The value of the equilibrium constant, Kc, for the reaction as written at 448 ^{o}C is 50.5. What are the concentrations of H_{2}, I_{2}, and HI in the flask at equilibrium? eqn:Kc=(HI+2x)^2/((H2x)(I2x)) Kc=50.5 Kc=50.5 HI=0 HI=0 x= Press [SOLVE]==> .x=.93498392612928 H2=1 H2=1 I2=2 I2=2 bound={1E99,1E99} bound={1E99,1E99} This gives us x, but what we really want are those equilibrium values:
eqn:Kc=(HI+2x)^2/((H2x)(I2x)) Kc=50.5 Kc=50.5 HI=0+2x HI=1.8699678522586 x= Press [SOLVE]==> .x=6.5557910050015E15 H2=1x H2=.06501607387072 I2=2x I2=1.06501607387072 bound={1E99,1E99} bound={1E99,1E99}Now x is essentially 0, which shows that we have the correct equilibrium values. (If the 'E15' is not showing on your calculator, press [2nd][>] to jump to the far end of the number to see what the exponent is.) Note that there are really two roots to this equation, but by some miraculous algorithm the TI calculators will always give you the smallest positive root, which is exactly what we want here! Answer: [HI] = 0+2x = 1.870 M, [H_{2}] = 1x = 0.06501 M, and [I_{2}] = 2x = 1.06501 M 
Additional Problems That Absolutely Require a Graphing Calculator
Ah, now the cool thing here is that we can do calculations that others can only dream of!
Without an equationsolving calculator, this sort of analysis is basically impossible.
For example, take Sample Exercise 15.10 and give it a twist:
For the Haber process, N_{2}(g) + 3 H_{2}(g) <==> 2 NH_{3}(g)
, K_{p} = 1.45E5 at 500 ^{o}C.
Starting with a partial pressure of H_{2} of 1.0 atm and a partial pressure of N_{2} of 2.0 atm, what are the
partial pressures at equilibrium?
eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=0 NH3=0 x= Press [SOLVE]==> .x=.0026586638154781 N2=2 N2=2 H2=1 H2=1 bound={1E99,1E99} bound={1E99,1E99}check: eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=0+2x NH3=.00531732763095 x= Press [SOLVE]==> .x=0 N2=2x N2=1.9973413361845 H2=13x H2=.99202400855357 bound={1E99,1E99} bound={1E99,1E99}Answer: [NH_{3}] = 0.00532 atm, [N_{2}] = 1.997 atm, and [H_{2}] = 0.9920 atm. OK, now what happens if we increase the partial pressure of NH_{3} to 2.0 atm? eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=2 NH3=2 x= Press [SOLVE]==> .x=.97456743083544 N2=1.9973413361845 N2=1.9973413361845 H2=.99202400855357 H2=.99202400855357 bound={1E99,1E99} bound={1E99,1E99}Answer: The reaction runs backward (x < 0)! Check: eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=2+2x NH3=0.0508651383291 x= Press [SOLVE]==> .x=1.4876168166883E14 N2=1.9973413361845x N2=2.9719087670199 H2=.992024008553573x H2=3.9157263010599 bound={1E99,1E99} bound={1E99,1E99} Final values: [NH_{3}] = 0.0509 atm, [N_{2}] = 2.972 atm, and [H_{2}] = 3.916 atm. Fine, now DOUBLE ALL THE PRESSURES (by reducing the volume). Let's see what happens: eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=0.0508651383291*2 NH3=.1017302766582 x= Press [SOLVE]==> .x=.047692266897154 N2=2.9719087670199*2 N2=5.9438175340398 H2=3.9157263010599*2 H2=7.8314526021198 bound={1E99,1E99} bound={1E99,1E99}Answer: Forward again, to reduce the total number of moles of gas. Check: eqn:Kp=(NH3+2x)^2/((N2x)(H23x)^3) Kp=1.4E5 Kp=1.4E5 NH3=.1017302766582+2x NH3=.19711481045251 x= Press [SOLVE]==> .x=2.549999999999E15 N2=5.9438175340398x N2=5.8961252671426 H2=7.83145260211983x H2=7.6883758014283 bound={1E99,1E99} bound={1E99,1E99} Final values: [NH_{3}] = 0.197 atm, [N_{2}] = 5.896 atm, and [H_{2}] = 7.688 atm.
