# What are algebraic steps can I use to solve this equation or get it into quadratic form neatly?

$$-15=40\tan(53) - 4.9\left(\frac{40}{v\cos(53)}\right)^2$$

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.

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Don't let the trig functions fool you-those are just constants. Your equation is $a=b-\frac c{v^2}$, so you can take this to $\frac c{v^2}=b-a$, invert both sides, multiply by $c$ and take the square root, remembering the $\pm$ sign. No need for the quadratic formula.