0=-16t^2+160t-144

Solution for 0=-16t^2+160t-144 equation:

0=-16t^2+160t-144

We move all terms to the left:

0-(-16t^2+160t-144)=0

We add all the numbers together, and all the variables

-(-16t^2+160t-144)=0

We get rid of parentheses

16t^2-160t+144=0

a = 16; b = -160; c = +144;
Δ = b2-4ac
Δ = -1602-4·16·144
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-128}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+128}{2*16}=\frac{288}{32}
=9
$