0=-16t^2+55t+10

Solution for 0=-16t^2+55t+10 equation:

0=-16t^2+55t+10

We move all terms to the left:

0-(-16t^2+55t+10)=0

We add all the numbers together, and all the variables

-(-16t^2+55t+10)=0

We get rid of parentheses

16t^2-55t-10=0

a = 16; b = -55; c = -10;
Δ = b2-4ac
Δ = -552-4·16·(-10)
Δ = 3665
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{-(-55)-\sqrt{3665}}{2*16}=\frac{55-\sqrt{3665}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+\sqrt{3665}}{2*16}=\frac{55+\sqrt{3665}}{32}
$