9=16t^2+160t

Solution for 9=16t^2+160t equation:

9=16t^2+160t

We move all terms to the left:

9-(16t^2+160t)=0

We get rid of parentheses

-16t^2-160t+9=0

a = -16; b = -160; c = +9;
Δ = b2-4ac
Δ = -1602-4·(-16)·9
Δ = 26176
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:

$\sqrt{\Delta}=\sqrt{26176}=\sqrt{64*409}=\sqrt{64}*\sqrt{409}=8\sqrt{409}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{409}}{2*-16}=\frac{160-8\sqrt{409}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{409}}{2*-16}=\frac{160+8\sqrt{409}}{-32}
$