can i solve this quadratic equation this way I was basically doing a physics problem and came across this equation in midway 
$\dfrac{2n-1}{n^2} = \dfrac{11}{36}$
then I equated $2n-1 = 11$ and $n^2 = 36$ and the value of $n$ which I got is $6$ in both the cases 
is this method valid? I mean I equated the ratio and ratio between many numbers are same 
I just want to know that whether I did it correctly or not.
 A: As a proportion: $\dfrac{A}{B} = \dfrac{C}{D} \to AD = BC$ we have:
$36(2n-1) = 11n^2$. So: $11n^2 - 72n + 36 = 0$, and $(11n-6)(n-6) = 0$. So $n = 6$ and $n = \dfrac{6}{11}$
A: No, it's not valid. For example, you can't solve $\dfrac{2n - 1}{n^2} = \dfrac{22}{36}$ in this way, but it's mathematically the same equation.
And before you think it has anything to do with the "simplest form" of the fraction on the RHS, consider $\dfrac{2n - 3}{n^2} = \dfrac{1}{3}$. The actual solution is $n = 3$, but by equating $2n - 3$ to $1$, you get $n = 2$.
A: It was by pure dumb luck that you were able to obtain a solution "by inspection".  Your solution is provably correct: if you substitute $n=6$, you do indeed get 11 for the numerator and 36 for the denominator.
However, the technique only worked in this case because of the exceptional coincidence that the same guess for $n$ happened to make everything work.  In other words, the technique doesn't generalize at all, and thus most people wouldn't have even noticed that a solution could be obtained by inspection.
Also, every quadratic has up to two real roots; I don't see any shortcut for obtaining the other real root.
