H(t)=-16t^2+256t+40

Solution for H(t)=-16t^2+256t+40 equation:

(H)=-16H^2+256H+40

We move all terms to the left:

(H)-(-16H^2+256H+40)=0

We get rid of parentheses

16H^2-256H+H-40=0

We add all the numbers together, and all the variables

16H^2-255H-40=0

a = 16; b = -255; c = -40;
Δ = b2-4ac
Δ = -2552-4·16·(-40)
Δ = 67585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-255)-\sqrt{67585}}{2*16}=\frac{255-\sqrt{67585}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-255)+\sqrt{67585}}{2*16}=\frac{255+\sqrt{67585}}{32}
$