# What are the zero(s) of: 9 = 15t - 4.9t^2?

Oct 24, 2015

$x \approx 0.82 , 2.24$

#### Explanation:

To use the quadratic formula, lets first re-arrange the equation.
$- 4.9 {t}^{2} + 15 t - 9$
$\frac{- 15 \pm \sqrt{{15}^{2} - 4 \left(- 4.9\right) \left(9\right)}}{2 \left(- 4.9\right)}$
(-15+-sqrt(48.6))/(2(-4.9)
So $x \approx 0.82 , 2.24$