110=-16t^2+98t+0.1

Solution for 110=-16t^2+98t+0.1 equation:

110=-16t^2+98t+0.1

We move all terms to the left:

110-(-16t^2+98t+0.1)=0

We get rid of parentheses

16t^2-98t-0.1+110=0

We add all the numbers together, and all the variables

16t^2-98t+109.9=0

a = 16; b = -98; c = +109.9;
Δ = b2-4ac
Δ = -982-4·16·109.9
Δ = 2570.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-\sqrt{2570.4}}{2*16}=\frac{98-\sqrt{2570.4}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+\sqrt{2570.4}}{2*16}=\frac{98+\sqrt{2570.4}}{32}
$