# rearrange $t - (m-q)^2 = v - (m-p)^2$ for quadratic formula form $ax^2 + bx +c = 0$ solving for $q$

I have the equation

$t - (m-q)^2 = v - (m-p)^2$

which I would like to rearrange to be able to apply the quadratic formula, and solve in terms of $q$. Accordingly, it needs to be in the form:

$ax^2 + bx +c = 0$

I have got as far as:

$(t-v)-(m-q)^2 + (m-p)^2 = 0$

$(t-v)-m^2 - q^2 + 2mq + m^2 +p^2 - 2mp = 0$

$(t-v) - q^2 + 2mq +p^2 - 2mp = 0$

But am now stuck.

Can anyone suggest a way forward? Or, as is likely, point out where I've gone wrong?

Thanks very much!

• What would play the role of $x?$ – mfl Jul 23 '14 at 11:49
• Apologies! I'm trying to solve in terms of $q$. I'll edit the question to reflect this. – user2728808 Jul 23 '14 at 11:51
• I think you are on the right track, for $x=q$, you have $a=-1$, $b=2m$, and $c=p^2-2mp+t-v$ – Tymric Jul 23 '14 at 11:59
• Thanks Timmy - your answer is just the same as kleineg. Thanks very much! – user2728808 Jul 23 '14 at 12:15

## 1 Answer

You have $(t-v) - q^2 + 2mq +p^2 - 2mp = 0$ and want to get the equation into the form $aq^2+bq+c$

I would combine the terms to get $- q^2 + 2mq + (p^2 - 2mp+t-v) = 0$

This gives you $a=-1$, $b=2m$, and $c =p^2 - 2mp+t-v$

• Many thanks, this is great! Sometimes it's hard to see what's right in front of you... – user2728808 Jul 23 '14 at 12:14
• Glad to be of help. – kleineg Jul 23 '14 at 12:28