NO. I was attempting to show a "general case" when the "C" = 0 (which is the case with your problem).

Let me do it another way:

Ax^2 + Bx + 0 = 0 : C=0; ok?

so: Ax^2 + Bx = 0

Now factor (no need for the quadratic):

x(Ax + B) = 0

so: x = 0 or Ax + B = 0 ; x = -B/A

If we use the quadratic:

Ax^2 + Bx + 0 = 0 ; then:

x = [-B +- sqrt(B^2 - 4*A*0)] / (2A)

x = [-B +- sqrt(B^2)] / (2A)

x = [-B +- B] / (2A)

x = (-B + B) / (2A) : so x = 0

or

x = (-B - B) / (2A) : so x = -2B / 2A = -B/A

So clearly the results are the same...so why use the quadratic ?