0=-16t^2-80t-29

Solution for 0=-16t^2-80t-29 equation:

0=-16t^2-80t-29

We move all terms to the left:

0-(-16t^2-80t-29)=0

We add all the numbers together, and all the variables

-(-16t^2-80t-29)=0

We get rid of parentheses

16t^2+80t+29=0

a = 16; b = 80; c = +29;
Δ = b2-4ac
Δ = 802-4·16·29
Δ = 4544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4544}=\sqrt{64*71}=\sqrt{64}*\sqrt{71}=8\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{71}}{2*16}=\frac{-80-8\sqrt{71}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{71}}{2*16}=\frac{-80+8\sqrt{71}}{32}
$