<s>[INST] Basic algebra, isolating the variable So I have the equation 
$$\tan30=\frac{4.9t-\frac{10}{t}}{\frac{8.77}{t}}$$
And I want to find t, but my algebra has failed me.
This is my working so far. 
$$\frac{8.77}{t}=\frac{4.9t-\frac{10}{t}}{\tan30}$$
$$\frac{8.77}{t}=\frac{4.9t}{\tan30}-\frac{10}{\tan30t}$$
$$8.77=\frac{4.9t}{\tan30t}-\frac{10}{\tan30t^2}$$
$$8.77=\frac{4.9}{\tan30}-\frac{10}{\tan30t^2}$$
$$0=\frac{4.9}{\tan30}-\frac{10}{\tan30t^2}-8.77$$
$$\frac{10}{\tan30t^2}=\frac{4.9}{\tan30}-8.77$$
Invert
$$\frac{\tan30t^2}{10}=\frac{\tan30}{4.9}-\frac{1}{8.77}$$
$$\frac{t^2}{10}=\frac{\tan30}{4.9\tan30}-\frac{1}{8.77\tan30}$$
$$t^2=\frac{10}{4.9}-\frac{10}{8.77\tan30}$$
$$t=\sqrt{0.065844}$$
$$=0.2566$$
However I know this is too long winded for the question, and the answer is wrong as well. So I am wondering 1- where I have gone wrong and 2- what is a better way of doing it. Thanks
 [/INST] Multiplying the right side of the equation by $\frac{t}{t} = 1$, with the assumption that $t \neq 0$, we have
$$\tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} \,\,.$$
Now, we can isolate $t$ as follows
$$\begin{align} \tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} &\implies 8.77\tan(30^{\circ}) = 4.9t^2 - 10 \\&\implies t^2 = \frac{8.77\tan(30^{\circ}) + 10}{4.9} \\&\implies t = \pm \sqrt{\frac{8.77\sqrt{3} + 10}{4.9}} \,\,. \end{align}$$
</s>