For all functions of the form f(x)=ax^2+bx+c which is true when b=0

I believe if b were to equal the value of 0, there would be 2 solutions to the function. Or not mainly if the resulting expression turns out to be a difference of squares.

Eg. f(x) = 2x^2 + 0x + 2

2x^2 + 2

2(x^2 + 1)

Here there would not be x intercepts, mainly because x cannot be solved.

1.17+3.99a. you combine the terms with a to create 3.99 and the 1.17 stays because it has no like terms or a variable