0=-16t^2+160t+25

Solution for 0=-16t^2+160t+25 equation:

0=-16t^2+160t+25

We move all terms to the left:

0-(-16t^2+160t+25)=0

We add all the numbers together, and all the variables

-(-16t^2+160t+25)=0

We get rid of parentheses

16t^2-160t-25=0

a = 16; b = -160; c = -25;
Δ = b2-4ac
Δ = -1602-4·16·(-25)
Δ = 27200
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:

$\sqrt{\Delta}=\sqrt{27200}=\sqrt{1600*17}=\sqrt{1600}*\sqrt{17}=40\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-40\sqrt{17}}{2*16}=\frac{160-40\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+40\sqrt{17}}{2*16}=\frac{160+40\sqrt{17}}{32}
$