Aprojectile is launched straight up from ground level with an initial velocity of 320 ft/sec when will it's height above ground be
1538 feet

Given: at time = 0, v0=+320 ft/s, [ assumed a=-g=-32.2 on earth ]

use kinematic equation for vertical projectiles,
Height,
H(t) = 1538 = v0(t)+(1/2)at^2=320t+(1/2)(32.2)t^2

Solve for t using the quadratic formula,
with A=16.1, B=320, C=-1538:
16.1t^2+320t-1538=0
t=8.14 or t=11.74

This means that at t=8.14, the projectile reaches 1538 feet (on its way up), and at t=11.74, the projectile falls back down and reaches also 1538 feet.

The answer would he -4,-2 because the points are just on the opposite side. since the x axis is +4 it's going to he -4 and since the y axis is -2 it stays there because whaen rotated it stays in the third quadrant