Hey there. I am having difficulties with the Quadratic formula,
I know what goes where in the cases, but its just solving from there.
like in this equasion.
2x^2-5x=0
I know how to get it into the quadratic formula
this is what I got.
Hey there. I am having difficulties with the Quadratic formula,
I know what goes where in the cases, but its just solving from there.
like in this equasion.
2x^2-5x=0
I know how to get it into the quadratic formula
this is what I got.
I'm surprised no one has pointed out that B is negative meaning that it should be
$\displaystyle \frac{5 \pm \sqrt{25}}{4}$
Since the quadratic is:
$\displaystyle \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
Making it
$\displaystyle -(-5)$ or just $\displaystyle 5 $
we would have gotten to that
we just had to deal with a more fundamental error first. anyway, so that we are all clear and on the same page:
the formula is: $\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
here $\displaystyle a = 2$, $\displaystyle b = -5$ and $\displaystyle c = 0$