2/3 = 5n-81/{([n-1]/4)^2 -10} What is n?
All help greatly appreciated
$\displaystyle
\frac{2}{3} = \frac{5n-81}{(\frac{(n-1)}{4})^2 -10}$
$\displaystyle
\frac{2}{3} = \frac{16(5n-81)}{(n-1)^2 -160}$
$\displaystyle
2(~(n-1)^2-160~) = 48(5n-81)$
$\displaystyle
(~(n)^2-159 -2n~) = 24(5n-81)$
Now use formula for roots of quadratic equation
Your formatting is ambiguous; I think you mean the following:
. . . . .$\displaystyle \frac{2}{3}\, =\, \frac{5n\, -\, 81}{\left(\frac{n\, -\, 1}{4}\right)^2\, -\, 10}$
If so, then a good first step in solving this rational equation is probably to multiply through, to get rid of the denominators:
. . . . .$\displaystyle 2\left[\left(\frac{n\, -\, 1}{4}\right)^2\, -\, 10\right]\, =\, 3(5n\, -\, 81)$
. . . . .$\displaystyle 2\left[\frac{n^2\, -\, 2n\, +\, 1}{16}\, -\, 10\right]\, =\, 15n\, -\, 243$
. . . . .$\displaystyle \frac{n^2\, -\, 2n\, +\, 1}{8}\, -\, 20\, =\, 15n\, -\, 243$
. . . . .$\displaystyle n^2\, -\, 2n\, +\, 1\, -\, 160\, =\, 120n\, -\, 1944$
Then collect all the terms on one side of the "equals" sign:
. . . . .$\displaystyle n^2\, -\, 122n\, +\, 1785\, =\, 0$
Then apply the Quadratic Formula to solve for "n". Remember to check your solutions in the original equation, making sure that they don't cause any "division by zero" problems.