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
 A: 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
$$\begin{align} \tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} &\implies 8.77\tan(30^{\circ}) = 4.9t^2 - 10 \\&\implies t^2 = \frac{8.77\tan(30^{\circ}) + 10}{4.9} \\&\implies t = \pm \sqrt{\frac{8.77\sqrt{3} + 10}{4.9}} \,\,. \end{align}$$
A: 1 -In the second line the t is dividing, hence it should be multiplying when you pass it to the other side of the equation in the bird line
2- In all cases you'll end up with a quadratic expression with a 0 in the linear term. You may find it easier to just add the 4.9t to the -10/t. You will get rid of the t coefficient right away.
A: Your algebra begins to break in your third step when you write $8.77 = {4.9\over{tan30t^2}} -{10\over{tan30t^2}}$.
An easier way to start for me is start by multiplying the whole right side by the reciprocal of the dividend (8.77/t). 
After this just multiply distributively and you will see a clearer way of getting to an equation that equals q. Have a look at how I solved it. Hopefully, you find it helpful. 
