0=-16t^2+92t-100

Solution for 0=-16t^2+92t-100 equation:

0=-16t^2+92t-100

We move all terms to the left:

0-(-16t^2+92t-100)=0

We add all the numbers together, and all the variables

-(-16t^2+92t-100)=0

We get rid of parentheses

16t^2-92t+100=0

a = 16; b = -92; c = +100;
Δ = b2-4ac
Δ = -922-4·16·100
Δ = 2064
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:

$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-4\sqrt{129}}{2*16}=\frac{92-4\sqrt{129}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+4\sqrt{129}}{2*16}=\frac{92+4\sqrt{129}}{32}
$