160+48t-16t^2=160

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

160+48t-16t^2=160

We move all terms to the left:

160+48t-16t^2-(160)=0

We add all the numbers together, and all the variables

-16t^2+48t=0

a = -16; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·(-16)·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*-16}=\frac{0}{-32}
=0
$