A shot leaves a thrower's hand 6.5 feet above the ground. The angle at which it was thrown is 45 degrees and it had a velocity of 44 $\displaystyle \frac{ft}{s}$. Find the time it hits the ground and the range.

The equations are as follows:

$\displaystyle x=x_o + (V_o cos\theta)t$ ; $\displaystyle y=y_o + (V_o tsin\theta - \frac{1}{2} gt^2)$

So basically what I have to do is, set the y equation equal to t and find a t value. Then plug that t value in for the x equation. Here's my work

$\displaystyle y=(\frac{-g}{2})t^2 + (V_o sin\theta)t + y_o$ which is in $\displaystyle (At^2 + Bt + C = 0)$ format.

$\displaystyle t=\frac{V_o sin\theta + \sqrt{(V_o sin\theta)^2 + 2gy_o}}{g}$

Subbed my values in and I got t=2.1348 s

Plugged that value for t in the x equation and got 66.4194 ft. How does my equation look for finding t? If its right then I think I can assume my values for t and x are correct as well.