160+48t-16t^2=0

Solution for 160+48t-16t^2=0 equation:

160+48t-16t^2=0

a = -16; b = 48; c = +160;
Δ = b2-4ac
Δ = 482-4·(-16)·160
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-112}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+112}{2*-16}=\frac{64}{-32}
=-2
$