0=-16t^2+60.8t+12.8

Solution for 0=-16t^2+60.8t+12.8 equation:

0=-16t^2+60.8t+12.8

We move all terms to the left:

0-(-16t^2+60.8t+12.8)=0

We add all the numbers together, and all the variables

-(-16t^2+60.8t+12.8)=0

We get rid of parentheses

16t^2-60.8t-12.8=0

a = 16; b = -60.8; c = -12.8;
Δ = b2-4ac
Δ = -60.82-4·16·(-12.8)
Δ = 4515.84
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{-(-60.8)-\sqrt{4515.84}}{2*16}=\frac{60.8-\sqrt{4515.84}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60.8)+\sqrt{4515.84}}{2*16}=\frac{60.8+\sqrt{4515.84}}{32}
$