0=-16t^2+260t-195

Solution for 0=-16t^2+260t-195 equation:

0=-16t^2+260t-195

We move all terms to the left:

0-(-16t^2+260t-195)=0

We add all the numbers together, and all the variables

-(-16t^2+260t-195)=0

We get rid of parentheses

16t^2-260t+195=0

a = 16; b = -260; c = +195;
Δ = b2-4ac
Δ = -2602-4·16·195
Δ = 55120
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:

$\sqrt{\Delta}=\sqrt{55120}=\sqrt{16*3445}=\sqrt{16}*\sqrt{3445}=4\sqrt{3445}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-260)-4\sqrt{3445}}{2*16}=\frac{260-4\sqrt{3445}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-260)+4\sqrt{3445}}{2*16}=\frac{260+4\sqrt{3445}}{32}
$