
Notice that the position equation has terms involving time and also time squared. Equations that involve a variable squared are called quadratic equations. An equation is said to be quadratic in x if x^{2} appears as the highest power of x in the equation. Quadratic equations are special for two reasons:
 There are special techniques for solving them.
 There are two solutions for the squared variable.


The variables c, b, and a in the math formula are the coefficients of x^{0} (= 1), x^{1} = x, and x^{2}.The two solutions are true for any values of the variables. In physics, we will sometimes need the quadratic solutions above; but in some cases we can also use the easier method of factoring a quadratic equation to solve for the two solutions.


To find the solution we start as usual by defining the initial position as x_{0} = 0. We are also given that the arrow comes back down to x_{0} = 0 again 5.0 s after it is fired upward. The position equation becomes a quadratic in time t, from which a t can be factored from each term to give this intermediate equation:
$$\begin{array}{ccc}0={v}_{0}t{\scriptscriptstyle \frac{1}{2}}g{t}^{2}& \text{\hspace{1em}}\text{\hspace{1em}}\to \text{\hspace{1em}}\text{\hspace{1em}}& 0=t\text{}\left({v}_{0}\frac{gt}{2}\right)\end{array}$$ 
There are two ways in which the result can be zero:
$$\begin{array}{cccc}\text{if}& t=0& \text{orif}& {v}_{0}\frac{gt}{2}=0\end{array}$$ 
The solution on the right gives us what we want, which is the value of v_{0} when t = 5.0 s. The solution on the left is also realistic since it says the arrow was on the ground (x = 0) at the start when t = 0.

 
