0=-16t^2-29t+6

Solution for 0=-16t^2-29t+6 equation:

0=-16t^2-29t+6

We move all terms to the left:

0-(-16t^2-29t+6)=0

We add all the numbers together, and all the variables

-(-16t^2-29t+6)=0

We get rid of parentheses

16t^2+29t-6=0

a = 16; b = 29; c = -6;
Δ = b2-4ac
Δ = 292-4·16·(-6)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-35}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+35}{2*16}=\frac{6}{32}
=3/16
$