H(t)=-16t^2+128t+40

Solution for H(t)=-16t^2+128t+40 equation:

(H)=-16H^2+128H+40

We move all terms to the left:

(H)-(-16H^2+128H+40)=0

We get rid of parentheses

16H^2-128H+H-40=0

We add all the numbers together, and all the variables

16H^2-127H-40=0

a = 16; b = -127; c = -40;
Δ = b2-4ac
Δ = -1272-4·16·(-40)
Δ = 18689
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{-(-127)-\sqrt{18689}}{2*16}=\frac{127-\sqrt{18689}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{18689}}{2*16}=\frac{127+\sqrt{18689}}{32}
$