k2=[H][SO4]/[HSO4]

.012=(.01+x)(x)/(.01-x)

Solve for x, and calculate pH where H is .01+x

When I am solving for x does it become

.012= .01x + x^2/ .01 -x.. what is the next step of solving for x?

Next step, multiply both sides by (.01-x)

.012(.01-x)=.01x+x^2

gather terms, put in standard binomial form, use the quadratic equation, and solve for x.

Your question actually bothers me. You are asking an algebra 1 question in order to solve a very complicated chemistry question. Are you in over your head in chem math? You may want to discuss this with an academic counselor.