H=-16t2+10t+5

Solution for H=-16t2+10t+5 equation:

=-16H^2+10H+5

We move all terms to the left:

-(-16H^2+10H+5)=0

We get rid of parentheses

16H^2-10H-5=0

a = 16; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·16·(-5)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{105}}{2*16}=\frac{10-2\sqrt{105}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{105}}{2*16}=\frac{10+2\sqrt{105}}{32}
$