0=-16t^2+160t+18

Solution for 0=-16t^2+160t+18 equation:

0=-16t^2+160t+18

We move all terms to the left:

0-(-16t^2+160t+18)=0

We add all the numbers together, and all the variables

-(-16t^2+160t+18)=0

We get rid of parentheses

16t^2-160t-18=0

a = 16; b = -160; c = -18;
Δ = b2-4ac
Δ = -1602-4·16·(-18)
Δ = 26752
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{26752}=\sqrt{64*418}=\sqrt{64}*\sqrt{418}=8\sqrt{418}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{418}}{2*16}=\frac{160-8\sqrt{418}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{418}}{2*16}=\frac{160+8\sqrt{418}}{32}
$