0=-16t^2-31t+31

Solution for 0=-16t^2-31t+31 equation:

0=-16t^2-31t+31

We move all terms to the left:

0-(-16t^2-31t+31)=0

We add all the numbers together, and all the variables

-(-16t^2-31t+31)=0

We get rid of parentheses

16t^2+31t-31=0

a = 16; b = 31; c = -31;
Δ = b2-4ac
Δ = 312-4·16·(-31)
Δ = 2945
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{-(31)-\sqrt{2945}}{2*16}=\frac{-31-\sqrt{2945}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{2945}}{2*16}=\frac{-31+\sqrt{2945}}{32}
$