\begin{equation} -15=40\tan(53) - 4.9\left(\frac{40}{v\cos(53)}\right)^2 \end{equation}
My Question:
What steps should I take algebraically to solve for positive $v$ neatly? Is the only way to solve this by getting it in quadratic form? If so, how can I do that without making it look too messy?
I found this equation by manipulating the kinematics equations in my physics course to solve for one of the variables $v_0$ in a particular problem.