H(t)=-16t^2+9t+40

Solution for H(t)=-16t^2+9t+40 equation:

(H)=-16H^2+9H+40

We move all terms to the left:

(H)-(-16H^2+9H+40)=0

We get rid of parentheses

16H^2-9H+H-40=0

We add all the numbers together, and all the variables

16H^2-8H-40=0

a = 16; b = -8; c = -40;
Δ = b2-4ac
Δ = -82-4·16·(-40)
Δ = 2624
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:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{41}}{2*16}=\frac{8-8\sqrt{41}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{41}}{2*16}=\frac{8+8\sqrt{41}}{32}
$