0=-16(t)^2+100(t)+29

Solution for 0=-16(t)^2+100(t)+29 equation:

0=-16(t)^2+100(t)+29

We move all terms to the left:

0-(-16(t)^2+100(t)+29)=0

We add all the numbers together, and all the variables

-(-16t^2+100t+29)=0

We get rid of parentheses

16t^2-100t-29=0

a = 16; b = -100; c = -29;
Δ = b2-4ac
Δ = -1002-4·16·(-29)
Δ = 11856
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:

$\sqrt{\Delta}=\sqrt{11856}=\sqrt{16*741}=\sqrt{16}*\sqrt{741}=4\sqrt{741}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-4\sqrt{741}}{2*16}=\frac{100-4\sqrt{741}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+4\sqrt{741}}{2*16}=\frac{100+4\sqrt{741}}{32}
$