algebra

h = -16t^2 + 137t + 100

148 = -16t^2 + 137t + 100

-16t^2+137t-48 = 0               Use quadratic formula to find the two values of   t

a = -16     b = 137     c = -48

$t = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$t = {-137 \pm \sqrt{137^2-4(-16)(-48)} \over 2(-16)}$             caculate away ! 

The hight of an object is given by h = -16t^2 + 137t + 100
Find the two points in time when the object is 148 feet above the ground . Round your answer to the nearest hundredth of a second (two decimal places)

Hello Guest!

$h = -16t ^ 2 + 137t + 100\\ 148= -16t ^ 2 + 137t + 100\\ -16t ^ 2 + 137t + 100-148=0$

$-16t ^ 2 + 137t -48=0$

$x = \large {-137 \pm \sqrt{137^2-4\cdot (-16)\cdot (-48)} \over 2\cdot(-16)}\\ x=\dfrac{-137\pm 125.288}{-32}$

$x_1=0.37\\ x_2=8.20$

The object is 0.37 seconds and 8.20 seconds after being shot

from a height of 100 feet at a height of 148 feet above ground.

  !