# How do you solve h = -16t^2 + 50t + 4 using the quadratic formula?

Apr 27, 2017

$h = \frac{25 + \sqrt{689}}{16}$

$h = \frac{25 - \sqrt{689}}{16}$

#### Explanation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where $a {x}^{2} + b x + c = 0$

$h = - 16 {t}^{2} + 50 t + 4$

Looking at $a {x}^{2} + b x + c = 0$
I can see that your values are...

$a = - 16$
$b = 50$
$c = 4$

Now just put those values into the quadratic formula

$h = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$h = \frac{- \left(50\right) \pm \sqrt{{\left(50\right)}^{2} - 4 \left(- 16\right) \left(4\right)}}{2 \left(- 16\right)}$

$h = \frac{- 50 \pm \sqrt{2500 + 256}}{- 32}$

Negative divided by negative makes our numerator and denominator positive

$h = \frac{50 \pm \sqrt{2756}}{32}$

You might think we are finished here but remember to always check if you can simplify the square root. In this case $2756$ can be divided by $4$, which will give us a $2$ outside of the square root.

$h = \frac{50 \pm \sqrt{4 \cdot 689}}{32}$

$h = \frac{50 \pm 2 \sqrt{689}}{32}$

Now notice how we can factor $2$ out of our problem, thanks to simplifying the square root

$h = \frac{2 \left(25 \pm \sqrt{689}\right)}{2 \left(16\right)}$

Cancel the common factors

$h = \frac{\cancel{2} \left(25 \pm \sqrt{689}\right)}{\cancel{2} \left(16\right)}$

$h = \frac{25 \pm \sqrt{689}}{16}$

$\textcolor{g r e e n}{h = \frac{25 + \sqrt{689}}{16}}$
$\textcolor{g r e e n}{h = \frac{25 - \sqrt{689}}{16}}$