Let f(x)=36x^2-12x+a^2-b^2=36x^2-12x+1+a^2-b^2-1=(6x-1)^2+(a-b)(a+b)-1. If a^2-b^2=1, the quadratic expression f(x)=(6x-1)^2. When f(x)=0, x=1/6. The general solution is (6x-1)^2=1+b^2-a^2, so 6x-1=+sqrt(1+b^2-a^2), making x=(1/6)(1+sqrt(1+b^2-a^2)). If b=any number, and a=+1, the square root disappears and x=(1/6)(1+b). Or just considering factorisation: f(x)=(6x-1)^2-b^2=(6x-1-b)(6x-1+b), and a=1.