I have to find with what value ''m'' those two equations are equivalent?

$\displaystyle x^2$ +$\displaystyle m^2$x + (6-5m)x = 0

and $\displaystyle x^2$ +2(m-3)x + $\displaystyle m^2$ -7m +12 = 0

I found that with m=3 ... I did this: $\displaystyle x^2$ +$\displaystyle m^2$x + (6-5m)x = $\displaystyle x^2$ +2(m-3)x + $\displaystyle m^2$ -7m +12

Not sure if this is the right way?