Question

(mx-md+b-c)^2 + (x-v)^2 -r^2 = 0

solve for x

Answer 1

(mx - md + b - c)^2 + (x - v)^2 - r^2 = 0

m^2x^2 -2m^2xd +2mxb - 2mxc + m^2d^2 - 2mdb + 2mdc +b^2 -2bc +c^2 + x^2 -2xv + v^2 - r^2 = 0

x^2(m^2+1) + x(-2m^2d+2mb-2mc-2v) + m^2d^2 - 2mdb + 2mdc +b^2 -2bc +c^2 + v^2 - r^2 = 0

a = m^2+1
b = -2m^2d+2mb-2mc+2v
c = m^2d^2 - 2mdb + 2mdc +b^2 -2bc +c^2 +v^2 - r^2

x = (-b±sqrt(b^2-4ac))/2a

right?

Answer 2

I mean b = -2m^2d+2mb-2mc-2v

Answer 3

Is "mx" concatenated variable?

Answer 4

no
it is m*x
and m*d

Answer 5

I'm using a software to solve, you can use it to check
Ans
\[\tiny x=\frac{-b m+c m+d m^2+v-\sqrt{-b^2+2 b c-c^2+2 b d m-2 c d m-d^2 m^2+r^2+m^2 r^2-2 b m v+2 c m v+2 d m^2 v-m^2 v^2}}{1+m^2}\]
\[\tiny x=\frac{-b m+c m+d m^2+v+\sqrt{-b^2+2 b c-c^2+2 b d m-2 c d m-d^2 m^2+r^2+m^2 r^2-2 b m v+2 c m v+2 d m^2 v-m^2 v^2}}{1+m^2}\]

Can you see it?

Answer 6

see the attachment 'capture'

Answer 7

yes I believe I got this right, thank you