SOLUTION: Doing a quadratic equation problem, i cam across a problem. the equation i got was x^2-6c-3=0. When plugging this problem into the equation, x=-b+/-(sqrt(b^2+4ac))/2a, I did not kn

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Doing a quadratic equation problem, i cam across a problem. the equation i got was x^2-6c-3=0. When plugging this problem into the equation, x=-b+/-(sqrt(b^2+4ac))/2a, I did not kn      Log On


   

Question 239988: Doing a quadratic equation problem, i cam across a problem. the equation i got was x^2-6c-3=0. When plugging this problem into the equation, x=-b+/-(sqrt(b^2+4ac))/2a, I did not know if you could plug the b value (-3) into the place of -b or if you had to make it a positive b.
Found 2 solutions by JimboP1977, solver91311:
Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
Well don't forget that the general form is ax%5E2%2Bbx%2Bc
So the value of b in your equation is -6 so (-1)(-6) = 6
so b = 6 in this instance.
Does that help?

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If is truly your equation, then if you compare it to the general form of a quadratic:



that , , and .

However, if that is the case, then there is no need for the quadratic formula because:







Although, if you insist on using the quadratic formula, you will find that:






reduces to the same answer.

If you actually meant , that is another story, and then I'm curious where you got the idea that .

John