# Solving a quadratic equation

• Nov 13th 2013, 10:59 AM
MathIsOhSoHard
Solving a quadratic equation
I have the following quadratic equation:
$\frac{1}{2}kd^2-\mu_kmgl-\frac{1}{2}mv_B^2=0$

I would like to solve for d.
So I wanna use this formula:
$x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}$

Now I know that here I have:
$a=\frac{1}{2}k$

However is it true to say that:
$b=0$
$c=-\mu_kmgl-\frac{1}{2}mv_B^2$

???? I'm wondering if c can include operators like plus and minus?
• Nov 13th 2013, 11:28 AM
SlipEternal
Re: Solving a quadratic equation
Yes, that is correct.
• Nov 13th 2013, 03:27 PM
vegasgunner
Re: Solving a quadratic equation
yes it can