H=-16t^2+40t+30

Solution for H=-16t^2+40t+30 equation:

=-16H^2+40H+30

We move all terms to the left:

-(-16H^2+40H+30)=0

We get rid of parentheses

16H^2-40H-30=0

a = 16; b = -40; c = -30;
Δ = b2-4ac
Δ = -402-4·16·(-30)
Δ = 3520
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:

$\sqrt{\Delta}=\sqrt{3520}=\sqrt{64*55}=\sqrt{64}*\sqrt{55}=8\sqrt{55}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{55}}{2*16}=\frac{40-8\sqrt{55}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{55}}{2*16}=\frac{40+8\sqrt{55}}{32}
$