H(3t)=-16t^2+40

Solution for H(3t)=-16t^2+40 equation:

(3H)=-16H^2+40

We move all terms to the left:

(3H)-(-16H^2+40)=0

We get rid of parentheses

16H^2+3H-40=0

a = 16; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·16·(-40)
Δ = 2569
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{-(3)-\sqrt{2569}}{2*16}=\frac{-3-\sqrt{2569}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{2569}}{2*16}=\frac{-3+\sqrt{2569}}{32}
$