H=-16t^2+29t+5

Solution for H=-16t^2+29t+5 equation:

=-16H^2+29H+5

We move all terms to the left:

-(-16H^2+29H+5)=0

We get rid of parentheses

16H^2-29H-5=0

a = 16; b = -29; c = -5;
Δ = b2-4ac
Δ = -292-4·16·(-5)
Δ = 1161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-3\sqrt{129}}{2*16}=\frac{29-3\sqrt{129}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+3\sqrt{129}}{2*16}=\frac{29+3\sqrt{129}}{32}
$