Question

the height of a punted football is approximated by the equation below.
y is the height (in feet) and x is the horizontal distance (in feet) from where the football was punted.
$y = -\frac{16}{2025}x^2 + \frac{9}{5}x + \frac{3}{2}$
use the numbers as they are shown desmos

1. how far from the punter does the football hit the ground?

(do not round answers)
use the numbers as they are shown desmos
type a response

Explanation:

Step1: Set height $y=0$

$0 = -\frac{16}{2025}x^2 + \frac{9}{5}x + \frac{3}{2}$

Step2: Multiply by 2025 to eliminate denominators

$0 = -16x^2 + 9 \times 405x + \frac{3}{2} \times 2025$
$0 = -16x^2 + 3645x + 3037.5$
Multiply by -1: $16x^2 - 3645x - 3037.5 = 0$

Step3: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here $a=16$, $b=-3645$, $c=-3037.5$
First calculate discriminant:
$\Delta = (-3645)^2 - 4 \times 16 \times (-3037.5)$
$\Delta = 13286025 + 194400 = 13480425$
$\sqrt{\Delta} = \sqrt{13480425} = 3671.57$ (exact value: $\sqrt{13480425} = \frac{15\sqrt{59913}}{2}$)
Then solve for $x$:
$x = \frac{3645 \pm 3671.57}{32}$
We discard the negative solution (distance can't be negative):
$x = \frac{3645 + 3671.57}{32} = \frac{7316.57}{32}$
Or using exact fractions:
$x = \frac{3645 + \sqrt{13480425}}{32} = \frac{7290 + 15\sqrt{59913}}{64}$
Simplify the decimal: $x = \frac{7290 + 15\sqrt{59913}}{64} = \frac{15(486 + \sqrt{59913})}{64}$
Calculating the exact numerical value without rounding:
$x = \frac{3645 + \sqrt{13480425}}{32} = \frac{3645 + 3671.5709}{32} = \frac{7316.5709}{32} = 228.642840625$

Answer:

$\frac{7290 + 15\sqrt{59913}}{64}$ or $228.642840625$ feet