How to do without quadratic formula? For the reaction #"CO"(g) + "NH"_3(g) rightleftharpoons "HCONH"_2(g)#, #K_c = 0.620#

I think you mean without using the quadratic formula. It depends on the problem, but maybe...

So, the method I often see is the "ICE" table method:

.........#CO+NH_3<=>HCONH_2#
I.....#"|"# 1.00.......2.00...........0.00
C...#"|"# -x............-2x............+x
E...#"|"# 1.00-x....2.00-2x.....x

When you write out the equilibrium expression, you get:

#K_c = 0.620 = ([HCONH_2])/([CO][NH_3])=(x)/((1-x)(2-x)) = x/(2-3x+x^2)#

#K_c = 0.620 = x/(x^2-3x+2)#
#0.620x^2-1.86x+1.24 = x#
#0.620x^2-2.86x+1.24 = 0#
#0.310x^2-1.43x+0.620 = 0#
#31.0x^2-143x+62.0 = 0#

You can't really do this without the quadratic formula. 143 does not share factors with 31 and 62 (its factors are 11 and 13, and not 31 or 2). So:

#(-(-1.43)pmsqrt((-1.43)^2-4*0.310*0.620))/(2*0.310)#
#=(1.43pmsqrt(~1.28))/(0.620)#
#=(1.43pm~1.13)/(0.620)#
#=4.13 "M"#
#=0.484 "M"#

Then you plug into an expression for one of the compounds, like #1.00-x#, and #0.484 "M"# works.

Checking the equilibrium expression:
#K_c = 0.620 = (0.484)/((1.00-0.484)(2.00-0.484)) ~~ 0.620#

...and yep, it will be #0.484 "M"# #HCONH_2# at equilibrium.