# How do you find the roots, real and imaginary, of  h=-16t^2+64t  using the quadratic formula?

Apr 1, 2018

0, 4

#### Explanation:

x=(-b±√b^2-4ac)/(2a)

In this case,
$a = - 16$
$b = 64$
$c = 0$

Plug in the numbers.

To make it simpler, let's only substitute $a$ first.

x=(-b±√b^2-(-64c))/-32

Substitute $b$.

x=(-64±√64^2+64c)/-32

Finally, substitute $c$.

x=(-64±√64^2)/-32

Simplify to get
x=(-64±64)/-32

If $x = \frac{- 64 + 64}{-} 32$, then $x = 0$ because $\frac{0}{-} 32 = 0$.

If $x = \frac{- 64 - 64}{-} 32$, then $x = 4$ because $- \frac{128}{-} 32 = 4$.

The roots of $h = - 16 {t}^{2} + 64 t$ are 0, 4.

Yay!